Electrostatics Notes

ELECTROSTATICS INTRODUCTION Electromagnetism is, almost unarguably, the most important basic technology in the world today. Almost every modern device, from cars to kitchen appliances to computers, is dependent upon it. Life, for msot of us, would be almost unimaginable without electromagetism. In fact, electromagnetism cuts such a wide path through modern life that the teaching of electromagnetism has developed into several different specialities. Initially electricity and magnetism were classified as independent phenomena, but different specialities. Initially electricity and magnetism were classified as independent phenomena, but after some experiments (we will discuss later) it was found they are interrelated so we use the name Electromagnetism. In electromagnetism we have to study basic properties of electromagnetic force and field (the term field will be introduced in later section). The electromagnetic force between charged particles is one of the fundamental force of nature. WE being this chapter by describing some of the basic properties of one manifestation of the electromagnetic force, the electrostatic force between charges (the force between two charges when they are at rest) under the heading electrostatics. Specific Properties of Charge (i) Charge is a scalar quantity : It represents excess or deficiency of electrons. (ii) Charge is transferable : If a charged body is put in contact with an another body, then charge can be transferred to another body. (iii) Charge is always associated with mass Charge cannot exist without mass though mass can exist without charge. (a) So the presence of charge itself is a convincing proof of existence of mass. (b) In charging, the mass of a body changes. (c) When body is given positive charge, its mass decreases. (d) When body is given negative charge, its mass increases. (iv) Charge is quantised The quantization of electric charge is the property by virtue of which all free charges are integral multiple of a basic unit of charge represented by e. Thus charge q of a body is always given by q = ne n = positive integer or negative integer The quantum of charge is the charge that an electron or proton carries. Note : Charge on a proton = (–) charge on an electron = 1.6 × 10–19 C (v) Charge is conserved In an isolated system, total charge does not change with time, though individual charge may change i.e. charge can neither be created nor destroyed. Conservation of charge is also found to hold good in all types of reactions either chemical (atomic) or nuclear. No exceptions to the rule have ever been found. (vi) Charge is invariant Charge is independent of frame of reference. i.e. charge on a body does not change whatever be its speed. (vii) Accelerated charge radiates energy (viii) Attraction – Repulsion Similar charges repel each other while dissimilar attract Methods of Charging (i) Friction If we rub one body with other body, electrons are transferred from one body to the other. Transfer of electrons takes places from lower work function body to higher work function body. Positive charge Negative charge Glass rod Silk cloth Woollen cloth Rubber shoes, Amber, Plastic objects Dry hair Comb Flannel or cat skin Ebonite rod Note : Clouds become charged by friction (ii) Electrostatic induction If a charged body is brought near a metallic neutral body, the charged body will attract opposite charge and repel similar charge present in the neutral body. As a result of this one side of the neutral body becomes negative while the other positive, this process is called “Electrostatic Induction”. Charging a body by induction (in four successive steps) charging body q'=0 charged body is brought near uncharged body step-1 charging body q'=-ve uncharged body is connected to earth step-2 charging body q'=-ve uncharged body is disconnected from the earth step-3 q'=-ve charging body is removed step-4 Some important facts associated with induction(i) Inducing body neither gains nor loses charge (ii) The nature of induced charge is always opposite to that of inducing charge (iii) Induction takes place only in bodies (either conducting or non conducting) and not in particles. (iii) (a) (b) (c) Conduction The process of transfer of charge by contact of two bodies is known as conduction. If a charged body is put in contact with uncharged body, the uncharged body becomes charged due to transfer of electrons from one body to the other. The charged body loses some of its charge (which is equal to the charge gained by the uncharged body) The charge gained by the uncharged body is always lesser than initial charge present on the charged body. Flow of charge depends upon the potential difference of both bodies. [No potential difference No conduction]. Positive charge flows from higher potential to lower potential, while negative charge flows from lower to higher potential. ELECTROSTATICS (iv) (v) (vi) (vii) Charge differs from mass in the following sense. (a) In SI units, charge is a derived physical quantity while mass is fundamental quantity. (b) Charge is always conserved but mass is not (Note : Mass can be converted into energy E=mc 2 (c) The quanta of charge is electronic charge while that of mass it is yet not clear. (d) For a moving charged body mass increases while charge remains constant. True test of electrification is repulsion and not attraction as attraction may also take place between a charged and an uncharged body and also between two similarly charged bodies. q For a non relativistic (i.e. v << c) charged particle, specific charge =constant m q For a relativistic charged particle decreases as v increases, where v is speed of charged body.. m Ex. When a piece of polythene is rubbed with wool, a charge of – 2 × 10–7 C is developed on polythene. What is the amount of mass, which is transferred to polythene. Sol. From Q = ne, So, the number of electrons transferred n  Q e = 2  1 0 7 1 .6  10 1 9 = 1.25 × 1012 Now mass of transferred electrons = n × mass of one electron = 1.25 × 10 12 × 9.1 × 10–31 = 11.38 × 10–19 kg Ex. Sol. 1012  – particles (Nuclei of helium) per second falls on a neutral sphere, calculate time in which sphere gets charged by 2C. Number of – particles falling in t second = 1012t Charge on – particle = +2e , So charge incident in time t = (1012t).(2e) Given charge is 2 C   2 × 10–6 = (1012t).(2e)  t  10 18 1.6  10 1 9  6.25 s Coulomb's Law The electrostatic force of interaction between two static point electric charges is directly proportional to the product of the charges, inversely proportional to the square of the distance between them and acts along the straight line joining the two charges. If two points charges q1 and q2 separated by a distance r. Let F be the electrostatic force between these two charges. According to Coulomb's law. F  q1 q2 and F  Fe = kq 1 q 2 r2 1 r + where – + q2 F 1 on 2 F2 on1 q1 2 q1 F2 on1 + F1 on 2 q2 2  1 9 Nm  k   9  10   k = coulomb's constant or electrostatic force constant 4  0 C2   Coulomb's law in vector form  F12 = force on q1 due to q2 =  F21 = F12 kq 1 q 2 r̂2 1 r2 q1 r12 kq 1 q 2 r̂12 (here r̂12 is unit vector from q1 to q2 ) r2 y Coulomb's law in terms of position vector  kq1 q 2 r r F12 = r r 3 (r1  r2 ) r1  r2 r1 O q1 r 21 r2 F21 r q2 x q2 Principle of superposition The force is a two body interaction, i.e., electrical force between two point charges is independent of presence or absence of other charges and so the principle of superposition is valid, i.e., force on a charged particle due to r r r number of point charges is the resultant of forces due to individual point charges, i.e., F1 = F12 + F13 +... Note : Nuclear force is many body interaction, so principle of superposition is not valid in case of nuclear force. When a number of charges are interacting, the total force on a given charge is vector sum of the forces exerted on it by all other charges individually F  kq 0 q1 kq 0 q 2 ... kq 0 q i kq q    2  ... 02 n in vector form F  kq 0 2 2 r1 r2 ri rn n qi r i 1 2 i ˆri Some important points regarding Coulomb’s law and electric force (i) The law is based on physical observations and is not logically derivable from any other concept. Experiments till today reveal its universal nature. The law is analogous to Newton’s law of gravitation : F = G (ii) (iii) (a) Electric force between charged particles is much stronger than gravitational force, i.e., FE >>FG. This is why when both FE and FG are present, we neglect FG. (b) Electric force can be attractive or repulsive while gravitational force is always attractive. (c) Electric force depends on the nature of medium between the charges while gravitational force does not. (d) The force is an action–reaction pair, i.e., the force which one charge exerts on the other is equal and opposite to the force which the other charge exerts on the first. The force is conservative, i.e., work done in moving a point charge once round a closed path under the action of Coulomb’s force is zero. The net Coulomb’s force on two charged particles in free space and in a medium filled upto infinity are F= (iv) (v) (vi) Ex. Sol. m1 m 2 with the difference that : r2  1 q1 q 2 1 q1 q 2 F and F' = . So = = K, 2 2  4  0 r 4  r F' 0 Dielectric constant (K) of a medium is numerically equal to the ratio of the force on two point charges in free space to that in the medium filled upto infinity. The law expresses the force between two point charges at rest. In applying it to the case of extended bodies of finite size care should be taken in assuming the whole charge of a body to be concentrated at its ‘centre’ as this is true only for spherical charged body, that too for external point. Although net electric force on both particles change in the presence of dielectric but force due to one charge particle on another charge particle does not depend on the medium between them. Electric force between two charges does not depend on neighbouring charges. If the distance between two equal point charges is doubled and their individual charges are also doubled, what would happen to the force between them? 1 qq F= ....(i) 4  0 r 2 q2 1 (2q)(2 q) 1 1 4 q2 Again, F' = or F' = = =F 4  0 (2r)2 4  0 r 2 4  0 4 r 2 So, the force will remain the same. ELECTROSTATICS Ex. Sol. A particle of mass m carrying charge '+q 1' is revolving around a fixed charge '–q 2' in a circular path of radius r. Calculate the period of revolution. 4 2mr 1 q1q2 2 = mr = 4 0 r 2 T2 T2 = ( 4 0 )r 2 ( 4 2mr ) q1q2 or  0mr q1q2 T = 4r  where r is the vector drawn from source charge is test charge. Ex. Sol. The force of repulsion between two point charges is F, when these are at a distance of 1 m. Now the point charges are replaced by spheres of radii 25 cm having the charge same as that of point charges. The distance between their centres is 1 m, then compare the force of repulsion in two cases. In 2nd case due to mutual repulsion, the effective distance between their centre of charges will be increased (d' > d) so force of repulsion decreases as F  1 d2 d d' Equilibrium of charged particles In equilibrium net electric force on every charged particle is zero. The equilibrium of a charged particle, under the action of Colombian forces alone can never be stable. (1) Equilibrium of three point charges x KQ 1 q KQ 2 q  0 (a) Two charges must be of like nature as Fq  x2  r  x 2 (b) Third charge should be of unlike nature as FQ1  Therefore x  (2) Q1 Q1  Q 2 r and q= Q1 KQ 1 Q 2 KQ 1 q  0 r2 x2 Q 1 Q 2 ( Q 1  Q 2 )2 Equilibrium of symmetric geometrical point charged system Value of Q at centre for which system to be in state of equilibrium q a q a a a q a Q Q q (i) For equilateral triangle Q = q 3 a q q (ii) For square Q = q a  q(2 2  1) 4 q r Q2 (3) Equilibrium of Suspended Point Charge System For equilibrium position Tcos  = mg and T sin  = Fe = Fe kQ 2  tan  = = mg x2 mg x2 kQ 2 (a) If  is small then tan 1 2 kQ  Q2l 3 x 2 kQ 2  x 3 x  ~ sin  =  =  x =    2 ~ mg 2 2l x mg  2  0 mg  (b) If whole set up is taken into an artificial satellite (geff  ~ 0) then T = Fe= kq 2 4l 2 Ex. For the system shown in figure find Q for which resultant force on q is zero. Sol. For force on q to be zero, charges q and Q must be of opposite of nature. Net attraction force on q due to charges Q = Repulsion force on q due to q 2 2 FA = FR Ex.  2 kq kQq =  q = 2 2 Q Hence q = – 2 2 Q ( 2a )2 a2 2 q q 180° q (0,a) a a (0,0) Q a F A Q (a,a) FA q FR Two identically charged spheres are suspended by strings of equal length. The strings make an angle of 30 0 with each other. When suspended in a liquid of density 0.8 g/cc the angle remains same. What is the dielectric constant of liquid. Density of sphere = 1.6 g/cc. 300 0 15 F mg Sol. When set up shown in figure is in air, we have tan 150 = F When set up is mg immersed in the medium as shown in figure, the electric force experienced by the ball will reduce and will be equal to 30 0 F and the effective gravitational force r will  l  become mg  1   Thus we have tan 150 = s   Ex. Sol. 15° F r F F 1   r  =2 mg     1 l mg r  1  l  s s   mg eff=mg 1 Given a cube with point charges q on each of its vertices. Calculate the force exerted on any of the charges due to rest of the 7 charges. The net force on particle A can be given by vector sum of force experienced by this particle due to all the other r kq q r r charges on vertices of the cube. For this we use vector form of coulomb's law F  r 1 r 2 3  r1  r2  r1  r2 ELECTROSTATICS   r kq 2  akˆ From the figure the different forces acting on A are given as FA1 = a3      Z      kq 2 aˆi  akˆ kq 2  aˆi  aˆj  akˆ r kq 2 ajˆ  akˆ r r , FA  ; FA  FA2  3 3 3 3 4 3a 2a 2a       (a,0,a) 3   r r kq 2 aˆi kq 2  aˆi  aˆj r kq 2  aˆj , , F  FA5  F  A6 A7 3 a3 a3 2a   1 4 X (a,a,a) (0,a,a) 2 A (0,0,0) 5 (a,0,0) 6 (0,a,0) (a,a,0) The net force experienced by A can be given as 7 y r r r r r r r r   1  kq 2  1   1  ˆi  ˆj  kˆ  Fnet  FA1  FA2  FA3  FA4  FA5  FA6  FA7   2  a 2  3 3    Ex. Five point charges, each of value +q are placed on five vertices of a regular hexagon of side Lm. What is the magnitude of the force on a point charge of value –q coulomb placed at the centre of the hexagon? Sol. If there had been a sixth charge +q at the remaining vertex of hexagon force due to all the six charges on –q at O will be zero (as the forces due to individual charges will balance each other). L E F D q q O q q -q A C q B r r Now if f is the force due to sixth charge and F due to remaining five charges. r r r rr r r rrr f f00 0  f  Ffff F FFf  F FFff FF  1 qq q2    4  0 L2 4  0 L2 Electric Field Electric field is the region around charged particle or charged body in which if another charge is placed, it experiences electrostatic force. r Electric field intensity E : Electric field intensity at a point is equal to the electrostatic force experienced by a unit positive point charge both in magnitude and direction.  If a test charge q 0 is placed at a point in an electric field and experiences a force F due to some charges (called ur uur F source charges), the electric field intensity at that point due to source charges is given by E  q0  If the E is to be determined practically then the test charge q 0 should be small otherwise it will affect the charge distribution on the source which is producing the electric field and hence modify the quantity which is measured. Ex. Sol. A positively charged ball hangs from a long silk thread. We wish to measure E at a point P in the same horizontal plane as that of the hanging charge. To do so, we put a positive test charge q 0 at the point and measure F/q 0. Will F/q0 be less than, equal to, or greater than E at the point in question? When we try to measure the electric field at point P then after placing the test charge at P, it repels the source charge (suspended charge) and the measured r Properties of electric field intensity E : (i) It is a vector quantity. Its direction is the same as the force experienced by positive charge. (ii) Direction of electric field due to positive charge is always away from it while due to negative charge, always towards it. (iii) Its S.. unit is Newton/Coulomb. (iv) Its dimensional formula is [MLT –3A–1] (v) Electric force on a charge q placed in a region of electric field at a point where the electric field  r r intensity is E is given by F  qE . Electric force on point charge is in the same direction of electric field on positive charge and in opposite direction on a negative charge. (vi) It obeys the superposition principle, that is, the field intensity at a point due to a system of charges is vector sum of the field intensities due to individual point charges. r r r r E  E1  E 2  E 3 + ..... i.e. (vii) Ex. Sol. Ex. Sol. It is produced by source charges. The electric field will be a fixed value at a point unless we change the distribution of source charges. Electrostatic force experienced by –3C charge placed at point 'P' due  to a system 'S' of fixed point charges as shown in figure is F  ( 21î  9 ĵ ) µN. (i) Find out electric field intensity at point P due to S. (ii) If now, 2C charge is placed and –3 C is removed at point P then force experienced by it will be. Q1 Q3 Q2 S Q4 (i) r r F  qE (ii) Since the source charges are not disturbed the electric field intensity at 'P' will remain same.  r ˆ = -3µC (E) (21iˆ  9j)µN P   N E = – 7 î – 3 ĵ C r r F 2C = +2( E  = 2(–7 î – 3 ĵ ) = (–14 î – 6 ĵ )  N Calculate the electric field intensity which would be just sufficient to balance the weight of a particle of charge –10 c and mass 10 mg. (take g = 10 ms2)  As force on a charge q in an electric field E is r r Fq = qE So, according to given problem: Fe [W = weight of particle] ur uur | F q |  | W| i.e., E= i.e., |q|E = mg mg = 10 N/C., in downward direction. |q| q A W E ELECTROSTATICS List of formula for Electric Field Intensity due to various types of charge distribution : Name / Type Formula Point charge Note * q is source charge. Kq E=  2  r̂ |r| *  r is vector drawn from source charge to the test point. outwards due to +charges & inwards due to –charges. Infinitely long line charge  2Kr̂ r̂ = 2 0r r Infinite non-conducting thin sheet Uniformly charged ring * is surface charge density. (assumed uniform) * n is unit normal vector. * x = distance of point on the axis from centre of the ring. * electric field is always along the axis.  n̂ 2 0 E R KQx 2  x2  3/2 Ecentre = 0 Infinitely large charged conducting sheet Uniformly charged hollow conducting/ nonconducting /solid conducting sphere  n̂ 0 (i) for r R  kQ E   2 r̂ |r| (ii) for r < R E=0 Uniformly charged solid nonconducting sphere (insulating material) (i) for r R  kQ E   2 r̂ |r | (ii) for r R  kQ E 3 r R * q is linear charge density (assumed uniform) * r is perpendicular distance of point from line charge. * r is radial unit vector drawn from the charge to test point. * Q is total charge of the ring * x = distance of point on the axis from centre of the ring. * electric field is always along the axis. * is the surface charge . density (assumed uniform) * is the unit vector perpendicular to the surface. *R is radius of the sphere. * r is vector drawn from centre of sphere to the point. * Sphere acts like a point charge. placed at centre for points outside the  sphere. * E is always along radial direction. * Q is total charge ( = 4 R2). ( = surface charge density)  *r is vector drawn from centre of sphere to the point * Sphere acts like a point charge placed at the centre for points outside the sphere  n *E is always along radial dir * Q is total charge ( 43  R3 ) . ( = volume charge density) * Inside the sphere E r. 2 * Outside the sphere E 1/r . Graph Ex. Find out electric field intensity at point A (0, 1m, 2m) due to a point charge –20C situated at point B( 2 m, 0, 1m). Sol. KQ r KQ E = r 3 r = r 2 r̂ |r| |r|  r = P.V. of A – P.V. of B  r |r| = = (- 2 î + ĵ + k̂ ) E= (P.V. = Position vector) ( 2)2  (1) 2  (1) 2 = 2 9 109  (20  106 ) (– 2 î + ĵ + k̂ ) = – 22.5 × 103 (– 2 î + ĵ + k̂ ) N/C. 8 Ex. Two point charges 2c and – 2c are placed at points A and B as shown in figure. Find out electric field intensity at points C and D. [All the distances are measured in meter]. Sol. Electric field at point C : (E A, EB are magnitudes only and arrows represent directions) Electric field due to positive charge is away from it while due to negative charge, it is towards the charge. It is clear that E B > EA . 2 C EA C –2 C X ENet = (EB – EA) towards negative X-axis  = K(2c) ( 2) 2  K(2c) (3 2) 2 towards negative X-axis = 8000 (– î ) N/C Electric field at point D : Since magnitude of charges are same and also AD = BD So, EA = EB So, Enet = 2EA cos =   Vertical components of E A and EB cancel each other while horizontal components are in the same direction. = Ex. EB K  10 6 2 = D A ( 2 ,0) EA EB 2EAcos B ( 2 ,0 ) 2.K(2c) cos45 0 22 9000 2 î N/C. Six equal point charges are placed at the corners of a regular hexagon of side ‘a’. Calculate electric field intensity at the centre of hexagon ? ELECTROSTATICS Ans Zero (By symmetry) The y-component of electric field due to all the elements will be cancelled out to each other. So net electric field intensity at the point will be only due to X-component of each element. Enet = Enet =  dE x =  dE Cos  = KQx [R  x 2 ]3/ 2 E will be max when kx dq 2 (R  x 2 )3/ 2 O r Graph for variation of E with r. 2 dE 2KQ R = 0, that is at x= and Emax = dx 3 3 R2 2 if x>>R, E = Case (ii) : if x<> y then a2 + y2 ~ a2  FE = 1 Qqy 4 0 a 3 x –q FE (Towards centre) Since, restoring force FE  y, therefore motion of charge the particle will be S.H.M. Time period of SHM: 1/ 2  163 0 ma 3  m m T = 2 = 2 =  Qq  Qq   k   3  4  a 0   Ex. Sol. Calculate electric field intensity at a point on the axis which is at distance x from the centre of half ring, having total charge Q distributed uniformly on it. The radius of half ring is R. Consider an element of small angle d at an angle  as shown. Coordinates of element : (R cos , R sin , 0) Coordinates of point : (0, 0, x) Now electric field due to element : ˆ uur K(Rd)  [R cos ˆi  R sin ˆj  xk] KR 2 cos d dE  0  Ex = dEx =   2 2 2 2 2 3/ 2 (R cos   R sin   x ) (R 2  x 2 )3/ 2 0  Ey = dEy =   0 KR 2 sin d 2KR 2 2KQR   2 2 3/ 2 2 2 3/ 2 (R  x ) (R  x ) (R 2  x 2 )3/ 2  KRxd KQx  2 2 2 3/ 2 (R  x ) (R  x 2 )3/ 2 0 Ez = dEz =  Enet = 2 x 2 y E E E KQ 2 (R  x 2 )3/ 2 = 2 z Alternate solution 4R 2  x2 2 y x dEy R  d   dq dE z dEx dEz Consider an element of charge dq at an angle  on circumference of half ring. Due to this element electric field at the point on axis, which is at a distance x from the centre of half ring is dE. This electric field can be resolved into three component dE EZ = EX = EY =  Ex. Sol. / 2  dE sin  sin  = 0 dEX / 2 / 2  dE cos  = /2 /2  KQx [R  x 2 ]3/ 2 / 2 dq = Rd, sin  = Kdq 2K R sin  ...........(2) 2 2 sin. cos x R 2  x2 R R 2 dEZ ...........(1) 2 dE sin  cos  = dEY R  x2  Enet = E 2X  E 2Y Derive the expression of electric field intensity at a point 'P' which is situated at a distance x on the axis of uniformly charged disc of radius R and surface charge density . Also, derive results for (i) x >> R (ii) x << R The disc can be considered to be a collection of large number of concentric rings. Consider an element of the shape of rings of radius r and of width dr. Electric field due to this ring at P is ELECTROSTATICS dE = K.2r.dr.x (r 2  x 2 )3/2 Put, r 2 + x 2 = y2 2rdr = 2ydy  dE = K.2y.dy.x = 2K.x ydy y3 y3 Electric field at P due to all rings is along the axis :- E   dE  E = 2Kx  R2  x2  x 1 dy = 2Kx. y2  1  y   x R2  x2    1  x 1 1    = 2Kx   x  = 2K R 2  x2   R2  x 2     x = 2  1  2 R  x2 0  If x >> R Cases : (i) E=  [ 1 2 0   , along the axis  x R2 x 2 1 x ]=  [1 – 2 0  R2  1  2  x   1/ 2 ] 2 = 1 R  [1 – 1 + + higher order terms] 2 x2 2 0 R 2 Q  R2 = 2 = 2 40 x 4  0 x 2 4 0 x i.e. behaviour of the disc is like a point charge. If x << R = (ii) E=   [1 – 0] = 20 2 0 i.e. behaviour of the disc is like infinite sheet. Ex. Sol. Calculate electric field at a point on axis, which at a distance x from centre of uniformly charged disc having surface charge density and R which also contains a concentric hole of radius r. Consider a ring of radius y (r < y < R) and width dy concentric with disc and in the plane of the disc. Due to this ring, the electric field at the point P : R K(dq)x dE = [X 2  Y 2 ]3/ 2 r K x .(2y)dy Enet =  [x 2  y 2 ]3/ 2 r x2  R 2 2  x r 2 x P R 2 kx Enet = 2 y [ dq = 2ydy] dt , put x2 + y2 = t, 2y. dy = dt t 3/ 2 dE =  1 1 x   2 2   2 2 20  x  r x R  away from centre Alternate method We can also use superposition principle to solve this problem. (i) Assume a disc without hole of radius R having surface charge density +  . (ii) Also assume a concentric disc of radius r in the same plane of first disc having charge density –  . Now using derived formula in last example the net electric field at the centre is :    E net = E R + Er =  1 1 x     away from centre. 2 2 2 0  r2  x 2 R x  Electric field due to uniformly charged wire (i) Line charge of finite length : Derivation of expression for intensity of electric field at a point due to line charge of finite size of uniform linear charge density . The perpendicular distance of the point from the line charge is r and lines joining ends of line charge distribution make angle 1 and 2 with the perpendicular line.  r 1 2 P Consider a small element dx on line charge distribution at distance x from point A (see fig.). The charge of this element will be dq = dx. Due to this charge (dq), the intensity of electric field at the point P is dE. Then dE = dE K(dq) K(dx) 2 2 = r x r2  x2 There will be two components of this field : Ex = dEX dEY Kdx .cos  2  x2  dE   dE cos    r x Assuming, x = r tan  dx = r sec2. d  1 So Ex = = K r sec 2   cos   d  r 2  r 2 tan 2  2 K r  1  cos .d = 2 K [sin1+ sin2] r Similarly y-component. K Ey = r  1  sin .d = 2 Net electric field at the point: Enet = E 2x  E 2y K [cos2 - cos1] r ...........(1) ELECTROSTATICS (ii) We can derive a result for infinitely long line charge: In above eq. (1) & (2), if we put 1 = 2 = 90º, we can get required result. E E Enet = E x = (iii) 2K r 1 r r For Semi- infinite wire : 1 = 90º and 2 = 0º, so, Ex = K K , Ey = r r Ex r Ey Ex. Sol. A point charge q is placed at a distance r from a very long charged thread of uniform linear charge density . Find out total electric force experienced by the line charge due to the point charge. (Neglect gravity). Force on charge q due to the thread,  2K  F=   .q  r  By Newton's  law, every action has equal and opposite reaction So, force on the thread = 2K .q r (away from point charge) Ex. Figure shows a long wire having uniform charge density  as shown in figure. Calculate electric field intensity at point P. Sol. 1 = 90º and 2 = 360º – 37º So Ex. Ex = K [sin1+ sin2] r Ey = K [cos2 - cos1] r Find electric field at point A, B, C, D due to infinitely long uniformly charged wire with linear charge density  and kept along z-axis (as shown in figure).Assume that all the parameters are in S.I. units. Sol. Ex. Sol. EA = 2K ˆ 2K ˆ (i)  E = ( j) B 3 4 EC = 2 K   3iˆ  4jˆ  2K  OC = 5  5  5   ED = 2 K   3iˆ  4jˆ    5  5  ED = EC A point charge q is placed at a distance r from a very long charged thread of uniform linear charge density . Find out total electric force experienced by the line charge due to the point charge. (Neglect gravity). Force on charge q due to the thread,  2K  F=   .q  r  By Newton's  law, every action has equal and 2K .q opposite reaction, so force on the thread = r (away from point charge) Electric field due to uniformly charged infinite sheet  Enet = 2 towards normal direction o Electric field due to An infinitely large, uniformly charged sheet Derivation of expression for intensity of electric field at a point which is at a perpendicular distance r from the thin sheet of large size having uniform surface charge density . Assume a thin strip of width dx at distance x from line AB (see figure), which can be considered as a infinite line charge of charge density  = dx Due to this line charge the electric field intensity at point P will be dE = K(dx) r2  x2 Take another element similar to the first element on the other side of AB. Due to symmetry, Y-component of all such elements will be cancelled out. ELECTROSTATICS So net electric field will be given by : Enet =  dE x  dE cos  =  = 2K(dx) r2  x2 × cos Assume, x = r tan  dx = r sec2. d  / 2  Enet = 2K Note:   / 2 r sec2  .d .cos  2 2 2 r  r tan    2o , away from sheet (1) The direction of electric field is always perpendicular to the sheet. (2) The magnitude of electric field is independent of distance from sheet. Ex. An infinitely large plate of surface charge density + is lying in horizontal xy plane. A particle having charge –qo and mass m is projected from the plate with velocity u making an angle  with sheet. Find : (i) The time taken by the particle to return on the plate.. (ii) Maximum height achieved by the particle. (iii) At what distance will it strike the plate (Neglect gravitational force on the particle) Sol.    Electric force acting on the particle Fe = qoE : Fe = (qo)   downward  2 o  So, acceleration of the particle : a = qo Fe = 2 m = uniform m o This acceleration will act like ‘g’ (acceleration due to gravity) So, the particle will perform projectile motion. (i) T= (ii) H= (iii) 2u sin  2u sin  = . g  qo      2 o m  u 2 sin 2  u 2 sin 2  = . 2g 2  qo     2o m  u 2 sin 2 u 2 sin 2 R= = . g  qo      2 o m  Ex. A block having mass m and charge Q is resting on a frictionless plane at a distance d from fixed large nonconducting infinite sheet of uniform charge density  as shown in Figure. Assuming that collision of the block with the sheet is perfectly elastic, find the time period of oscillatory motion of the block. Is it SHM ? Sol. The situation is shown in Figure. Electric force produced by sheet will accelerate the block towards the sheet producing an acceleration. Acceleration will be uniform because electric field E due to the sheet is uniform. F QE = , where E = /20 m m As initially the block is at rest and acceleration is constant, from second equation of motion, time taken by the block to reach the wall a= d= 1 2 at 2 i.e., t= 2md = QE 2L = a E (– ) QE d 4md0 Q As collision with the wall is perfectly elastic, the block will rebound with same speed and as now its motion is opposite to the acceleration, it will come to rest after travelling same distance d in same time t. After stopping, it will again be accelerated towards the wall and so the block will execute oscillatory motion with 'span' d and time period. T = 2t = 2 4md0 2md =2 QE Q However, as the restoring force F = QE is constant and not proportional to displacement x, the motion is not simple harmonic. Ex. If an isolated infinite sheet contains charge Q1 on its one surface and charge Q2 on its other surface then prove that electric field intensity at a point in front of sheet will be Sol. Electric field at point P :  r r E = E Q1  E Q2 = Q2 Q1 n̂ n̂ + 2A 0 2A 0 = Q1  Q 2 n̂ 2A0 = Q , where Q = Q1 + Q2 2AO Q n̂ 2A 0 [This shows that the resultant field due to a sheet depends only on the total charge of the sheet and not on the distribution of charge on individual surfaces]. Ex. Three large conducting parallel sheets are placed at a finite distance from each other as shown in figure. Find out electric field intensity at points A, B, C & D. y x Sol. For point A Q  Q 3Q 2Q r r r E net = E Q + E 3Q + E –2Q = – 2A î – 2A î + 2A î =  A î 0 0 0 0 ELECTROSTATICS For point B: 3Q 2Q Q r r r r E net = E 3Q + E –2Q + E Q = – 2A î + 2A î + 2A î = 0 0 0 0 For point C : 2Q 2Q Q 3Q r r r r E net = E Q + E 3Q + E –2Q = + 2A î – 2A î – 2A î = – A î 0 0 0 0 For point D : Q Q 3Q 2Q r r r r E net = E Q + E 3Q + E –2Q = + 2A î + 2A î – 2A î = A î 0 0 0 0 Ex. Determine and draw the graph of electric field due to infinitely large non-conducting sheet of thickness t and uniform volume charge density  as a function of distance x from its symmetry plane. (a) x  Sol. t 2 (b) x  t 2 t  t  We can consider two sheets of thickness   x  and   x  . 2   2  Where the point P lies inside the sheet. Now, net electric field at point P : E = E1 – E2 = Q1 Q2 – 2A 0 2A 0 [Q1 : charge of left sheet; Q2 : charge of right sheet.] t  t  A   x   A   x  x 2  2  = = 2A0 0 For point which lies outside the sheet we can consider a complete sheet of thickness t t Q 2A0 E= tA t = 2A0 2 0 Alternate We can assume thick sheet to be made of large number of uniformly charged thin sheets. Consider an elementary thin sheet of width dx at a distance x from symmetry plane. Charge in sheet = Adx x (A : assumed area of sheet) dx Surface charge density, s= Adx A so, electric field intensity due to elementary sheet : dx dE = 2 0 (a) When x < t  2 (b) When x > t 2 x ENet = dx –  2 0 t /2 t/2  x x dx =  2 0 0 t/2 ENet = Ex. dx t = 2 20 0  t /2  Thin infinite sheet of width w contains uniform charge distribution . Find out electric field intensity at following points : (a) A point which lies in the same plane at a distance d from one of its edge. (b) A point which is on the symmetry plane of sheet at a perpendicular distance d from it. Sol. (a) Consider a thin strip of width dx. Linear charge density of strip :  = dx So, electric field due to this strip at point P dE = Enet =  dw d  dx  dw ln  20 x = 20  d  2kdx x ELECTROSTATICS (b) Consider a thin strip of width dx. Linear charge density of strip :  = dx  Ep = or Ep = 2. 0  2dEcos w /2 = d  0  w/2 0 dx 2 20 d  x dx = d2  x2 2 . d 2 d  x2  w tan–1  0 2d Electric field due to uniformly charged spherical shell E= KQ r2 E=0 r  R  For the outside points & point on the surface the uniformly r R rR  Rr 2 R 3    Rz   z  z r  R ELECTROSTATICS For The Points Inside The Sphere : Now lets derive the electric field due to uniformly charged solid sphere at a point 'P' inside it. The sphere is divided into two parts, the rings on the left part of point 'P' will produce electric field towards right and the rings on right part will produce electric field towards left and E net = E right – Eleft. For this, limit of integration is divided into two parts. E net   r  cos1   R   0  Electric      field due   to rings      1  r   of right   cos  R   part    As z2 = R2 + r 2 – 2rR cos When  = cos–1   z= When =0  z=R–r When =  z=R+r r  R Right Left P  Electric     field due   to rings     of left   part    R2  r2 From the result of previous case and just by changing limits we can write R 2  r2  Rr 2 R 3    Rz  Ein =   z z  R r  Rr 2 R 3    Rz  –  z z   R r R 2  r2 On solving this expression, we will get E and E = 0 if r < R. Finding electric field due to shell by integration is very lengthy, so we will not use this method. The given hand-out was just for knowledge. The best method to find E due to shell is by Gauss theorem which we will study later. Ex. Sol. Figure shows a uniformly charged sphere of radius R and total charge Q. A point charge q is situated outside the sphere at a distance r from centre of sphere. Find out the following : (i) Force acting on the point charge q due to the sphere. (ii) Force acting on the sphere due to the point charge. (i) Electric field at the position of point charge r KQ E  2 rˆ r r r KqQ KqQ | F|  2 F  2 rˆ  r r (ii) Since we know that every action has equal and opposite reaction so So, KqQ  F sphere =  r 2 r̂  KqQ  | F sphere| = 2 . r Q q Ex. Figure shows a uniformly charged thin sphere of total charge Q and radius R. A point charge q is also situated at the centre of the sphere. Find out the following : A B q C (i) Force on charge q (ii) Electric field intensity at A. (iii) Electric field intensity at B. Sol. (i) Electric field at the centre of the uniformly charged hollow sphere = 0 So force on charge q = 0 (ii) Electric field at A r r r Kq E A = E Sphere  E q = 0 + 2 ; r = CA r E due to sphere = 0 , because point lies inside the charged hollow sphere.    (iii) Electric field EB at point B = E Sphere  E q = K(Q  q) KQ Kq .rˆ ; r = CB .rˆ  2 .rˆ = r2 r r2 Note : Here, we can also assume that the total charge of sphere is concentrated at the centre, for calculation of electric field at B. Ex. Sol. Two concentric uniformly charged spherical shells of radius R1 and R2 (R2 > R1) have total charges Q1 and Q2 respectively. Derive an expression of electric field as a function of r for following positions. (i) r < R1 (ii) R1  r < R2 (iii) r  R2 (i) For (ii) therefore, point lies inside both the spheres Enet = EInner + Eouter = 0 + 0 For R1  r < R2, point lies outside inner sphere but inside outer sphere:  Enet = Einner + Eouter Q1 Q2 R1 R2 r < R1, KQ1 KQ1 r̂ r2 r (iii) For r  R2 point lies outside inner as well as outer sphere. Therefore, ENet = Einner + Eouter = = Ex. (i) 2 r̂ + 0 = KQ KQ1 r̂  2 2 r̂ 2 r r = K(Q1  Q 2 ) r̂ r2 A spherical shell having charge +Q (uniformly distributed) and a point charge + q0 are placed as shown. Find the force between shell and the point charge(r>>R). Force on the point charge + q0 due to the  shell = q0 E shell KQq 0  KQ  r̂ = (q0)  2  r̂ = r2  r  +Q , R r O +q0 P ELECTROSTATICS where r̂ , is unit vector along OP.. From action - reaction principle, force on the shell due to the point charge will also be : Fshell = KQq 0 ˆ ( r) r2 Conclusion :- To find the force on a hollow sphere due to outside charges , we can replace the sphere by a point charge kept at centre. Ex. Sol. Find force acting between two shells of radius R1 and R2 which have uniformly distributed charges Q1 and Q2 respectively and distance between their centres is r. + Q1, R1 + Q2, R2 r The shells can be replaced by point charges kept at centre, so force between them + Q1, R1 F= + Q2, R2 r KQ1Q2 ; r2 r F +Q1 +Q2 F Electric field due to uniformly charged solid sphere Derive an expression for electric field due to solid sphere of radius R and total charge Q which is uniformly distributed in the volume, at a point which is at a distance r from centre for given two cases. (i) r  R (ii) r  R Assume an elementary concentric shell of charge dq. Due to this shell, the electric field at the point (r > R) will be: dE = Kdq r2 Enet =  dE dq [from above result of hollow sphere] = KQ r2 For r < R, there will be no electric field due to shell of radius greater than r, so electric field at the point will be present only due to shells having radius less than r. E´net = Here, Q' =  KQ ' r2 4 Qr 3  r 3 = 3 4 3 3 R R 3 E´net = Q KQ´ KQr  3 ; r2 R away from the centre. NOTE : The electric field inside and outside the sphere is always in radial direction. Ex. Sol. A solid non conducting sphere of radius R and uniform volume charge density  has its centre at origin. Find out electric field intensity in vector form at following positions : (ii)  R R  , ,0   2 2  (i) (R/2, 0, 0) (i) At (R/2, 0, 0) : Distance of point from centre = the sphere, so (iii) (R, R, 0) (R / 2)2  02  02 = R/2 < R, so point lies inside r  pr R E = 3 = 3 [ î ] 2 0 0 (ii) (iii)  R R  , , 0  ; distance of point from centre = (R / 2) 2  (R / 2)2  02 = R = R, so point lies at At   2 2  the surface of sphere, therefore 4 K R 3 r  R ˆ R ˆ   R ˆ R ˆ KQ r 3 i j = i j E = 3 r =  3 R 3 0  2 2  R  2 2  The point is outside the sphere So, Ex. 4 K R 3   r KQ r 3  Riˆ  Rjˆ  =  Riˆ  Rjˆ  E = 3 r =    3 6 2 0  r ( 2R) A Uniformly charged solid non-conducting sphere of uniform volume charge density  and radius R is having a concentric spherical cavity of radius r. Find out electric field intensity at following points, as shown in the figure : Sol. (i) Point A (ii) Point B (iii) Point C (iv) Centre of the sphere R r O A B C Method I : (i) For point A : We can consider the solid part of sphere to be made of large number of spherical shells which have uniformly distributed charge on its surface. Now, since point A lies inside all spherical shells so electric field intensity due to all shells will be zero. EA = 0 (ii) For point B : All the spherical shells for which point B lies inside will make electric field zero at point B. So electric field will be due to charge present from radius r to OB. So, (iii) EB 4 K (OB3  r 3 ) uuur  [OB3  r 3 ] uuur = 3 = OB OB 30 OB3 OB3 For point C, similarly we can say that for all the shell points C lies outside the shell So, EC = K[ 43 (R  r 3 )] uuur  R 3  r3 OC = 30 [OC]3 [OC]3 uuur OC ELECTROSTATICS Method : II We can consider that the spherical cavity is filled with charge density  and also –, thereby making net charge density zero after combining. We can consider two concentric solid spheres: One of radius R and charge density  and other of radius r and charge density –. Applying superposition principle : – + uuur (i)  r uuur uuur uur uuur (OA) [(OA)] + =0 E A = E + E  = 30 30 4  uuur K  r 3 ()  uuur  uuur uur (OB) r 3  uuur 3  OB    OB = = E  + E  = + =  3 3  3 0 (OB) 30  30 3 0 (OB)  (ii) EB (iii) uur uur uuur E C = E + E  (iv) uuur uur uuur EO = E  + E  4  4  K  r 3 ()  K  R 3  uuur 3  uuur  OC +  3 =  OC 3 3 OC OC =  r 3  uuuur 1  3  OB  OB  uuur  3 3 [R  r ] OC 30 (OC)3 = 0 + 0 =0 Ex. In above question, if cavity is not concentric and centered at point P then repeat all the steps. Sol. Again assume  and – in the cavity, (similar to the previous example) : (i) uuur uuur uuur uur uuur [OA] ()PA + E A = E  + E  = 3 30 0 = Note : P A O B C  uuur uuur  uuur [OP] [ OA – PA ] = 30 30 Here, we can see that the electric field intensity at point A is independent of position of point A inside the cavity. Also the electric field is along the line joining the centres of the sphere and the spherical cavity. (ii) uur uuur uur E B = E  + E  uuur (OB) K[ 43 r 3 ( )] uuur = + PB 3 0 [PB]3 (iii) uur uuur uur EC = E  + E  K[ 43 R 3] uuur K[ 43 r 3 ()] uuur PC OC + = [OC]3 [PC]3 (iv) uuur uuur uur EO = E  + E  K[ 43 r 3 ()] uuur PO =0 + [PO]3 Ex. A non-conducting solid sphere has volume charge density that varies as  = 0 r, where 0 is a constant and r is distance from centre. Find out electric field intensities at following positions. (i) r R (ii) r < R Assume a ring of width Rd at angle  from X axis (as shown in figure). Potential due to the ring at the point P will be dV = K(dq) (r  R cos ) 2  (R sin ) 2 Where dq = 2R sin (Rd) where Q, = 4R2 then, net potential KQ 2 Solving this eq.we find V =  dV  &   0 V= KQ (for r > R) r V= KQ for (r < R) R sin .d (r  R cos ) 2  (R sin ) 2 Alternate Method : As the formula of E is easy , we use V = – (i) At outside point (r > R): rr  r rr Vout = – K Q  2  dr r  r   r r E·dr KQ Vout = r   For outside point, the hollow sphere acts like a point charge. (ii) Potential at the centre of the sphere (r=0) : As all the charges are at a distance R from the centre , So, Vcentre = (iii) KQ  R KQ  Radius of the sphere  Potential at inside point ( r < R ) : Suppose we want to find potential at point P , inside the sphere.  Potential difference between Point P and O : P r r – VP - VO =  Ein ·dr Where, Ein = 0 O So V P - VO = 0  VP = VO =  Vin = KQ R KQ  R KQ  Radius of the sphere  KQ  Dis tan ce from centre  (5) POTENTIAL DUE TO UNIFORMLY CHARGED SOLID SPHERE Derivation of expression for potential due to uniformly charged solid sphere of radius R and total charge Q (distributed in volume), at a point which is at a distance r from centre for the following situations. (i) r  R (ii) r  R Consider an elementary shell of radius x and width dx (i) For r  R : R K  4x 2 dx KQ = r r 0 V (ii) dx for r  R r x R K  4x 2 dx K4x 2 dx V  r x 0 r = KQ (3R2 – r2) 2R 3  From definition of potential (i) For r  R :   Q   4  R 3 3       r KQ KQ r̂  dr  2 r r  V =  (ii) For r  R : r KQr KQ V = –  2 .dr –  3 dr R r R  R V= Ex. KQ KQ 2 – [r – R2] R 2R 3 = KQ KQ 2 2 2 (3R2 – r2) 3 [2R – r + R ] = 2R 2R 3 Two concentric spherical shells of radius R1 and R2 (R2 > R1) are having uniformly distributed charges Q1 and Q2 respectively. Find out potential C (i) at point A R B (ii) at surface of smaller shell (i.e. at point B) AR (iii) at surface of larger shell (i.e. at point C) Q2 Q1 (iv) at r  R1 (v) at R1  r  R2 (vi) at r  R2 2 1 Sol. Using the results of hollow sphere as given in the table 7.4. KQ 2 KQ1 + R2 R1 (i) VA = KQ1 KQ 2 + R1 R2 (ii) VB = (iii) VC = KQ 2 KQ1 + R2 R2 (iv) for r  R1 , (v) for R1  r  R2 , V= KQ 2 KQ1 + R2 r (vi) for r  R2 , V= KQ1 KQ 2 + r r V= KQ 2 KQ1 + R2 R1 ELECTROSTATICS Ex. Two hollow concentric non-conducting spheres of radius the a and b (a > b) contain charges Qa and Qb respectively. Prove that potential difference between the two spheres is independent of charge on outer sphere. If outer sphere is given an extra charge, is there any change in potential difference? Sol. Vinner sphere = Vouter sphere = KQ b KQ a + b a Qb KQ b KQ a + a a Vinner sphere – V outer sphere = b Qa a KQ b KQ b – b a 1 1 V = KQb    b a    Which is independent of charge on outer sphere. If outer sphere in given any extra charge, then there will be no change in potential difference. Potential difference The potential difference between two points A and B is work done by external agent against electric field in taking a unit positive charge from A to B with no change in kinetic energy between initial and final points ie. K = 0 or Ki = K f (a) Mathematical representation If (W A  B)ext = Work done by external agent against electric field in taking the unit charge from A to B Then, Note : (b) A q B r r =   E.dr A Take W and q both with sign Properties (i) (ii) (iii) (iv) (1) UB  U A (WA  B )electric  (WA  B )ext   VB – V A =  = = = q q q   K  0 B r r   Fe .dr The difference of potential between two points is called potential difference. It is also called voltage. Potential difference is a scalar quantity. Its S.I. unit is also volt. If VA and VB be the potential of two points A and B, then work done by an external agent in taking the charge q from A to B is (Wext)AB= q (VB – VA) or (Wel) AB = q (VA – VB) . Potential difference between two points is independent of reference point. Potential difference in a uniform electric field  VB – V A = – E  AB  VB – VA = – |E| |AB| cos  = – |E| d = – Ed where d = effective distance between A and B along electric field. or we can also say that E = V d B A d E d Special Cases : Case 1. A B E Line AB is parallel to electric field.  Case 2. VA – VB = Ed Line AB is perpendicular to electric field.  VA – VB = 0  A VA = VB d E B Note : In the direction of electric field potential always decreases. Ex. 1C charge is shifted from A to B and it is found that work done by an external force is 40J in doing so against electrostatic forces, then find potential difference VA – VB Sol. (WAB)ext = q(VB – VA)  40 J = 1C (VB – VA)  VA – VB = – 40 V Ex. A uniform electric field is present in the positive x-direction. f the intensity of the field is 5N/C then find the potential difference (VB –VA) between two points A (0m, 2 m) and B (5 m ,3 m) Sol. VB – VA = – E . AB r  = – î (5) . (5 î + ĵ ) = –25V.. The electric field intensity in uniform electric field, E = V d Where V = potential difference between two points. d = effective distance between the two points. (projection of the displacement along the direction of electric field.) Ex. Find out following (i) VA – VB (ii) VB – VC (iii) VC – VA (iv) VD – VC (v) VA – VD (vi) Arrange the order of potential for points A, B, C and D. Sol. (i) VAB  Ed = 20 × 2 × 10–2 = 0.4 So, VA – VB = 0.4 V because In the direction of electric field potential always decreases. (ii) VBC  Ed = 20 × 2 × 10–2 = 0.4 so, (iii) VCA  Ed = 20 × 4 × 10–2 = 0.8 so, VC – VA = – 0.8 V VB – VC = 0.4 V because In the direction of electric field potential always decreases. (iv) VDC  Ed = 20 × 0 = 0 so, VD – V C = 0 because the effective distance between D and C is zero. (v) (vi) VAD  Ed = 20 × 4 × 10–2 = 0.8 so, VA – VD = 0.8 V because In the direction of electric field potential always decreases. The order of potential is : VA > VB > VC = VD . ELECTROSTATICS Potential difference due to infinitely long wire Derivation of expression for potential difference between two points, which have perpendicular distance rA and r B from infinitely long line charge of uniform linear charge density  : rA From definition of potential difference : A rB rB 2 K uur r uur   VAB = VB – VA =   E  dr  r r̂  dr B rB rA rA r   VAB = –2K n  B   rA  Potential difference due to infinitely long thin sheet Derivation of expression for potential difference between two points, having separation d in the direction perpendicularly to a very large uniformly charged thin sheet of uniform surface charge density . Let the points A and B have perpendicular distance r A and rB respectively then from definition of potential difference. r rB B     r̂  dr  E  dr VAB = VB – VA = 2 0  rA  VAB = –  rA  d (r – r ) = – 2 o B A 2 o Equipotential Surface Definition : If potential of a surface (imaginary or physically existing) is same throughout, then such surface is known as an equipotential surface. Properties of equipotential surfaces (i) When a charge is shifted from one point to another point on an equipotential surface, then work done against electrostatic forces is zero. (ii) Electric field is always perpendicular to equipotential surfaces. (iii) Two equipotential surfaces do not cross each other. Examples of equipotential surfaces (i) Point charge : Equipotential surfaces are concentric and spherical as shown in figure. In figure, we can see that sphere of radius R1 has potential V1 throughout its surface and similarly for other concentric sphere potential is same. V1 q V2 R1 R2 (ii) Line charge : V2 Equipotential surfaces have curved surfaces as that of V1 coaxial cylinders of different radii. (iii) Uniformly charged large conducting / non conducting sheets: Equipotential surfaces are parallel planes. + + + + + + + + + V1 V2 V3 Note : In uniform electric field equipotential surfaces are always parallel planes. Ex. Some equipotential surfaces are shown in figure. What can you say about the magnitude and the direction of the electric field ? y(cm) V V 0 V 60 V 20 0 10 2 1 28 37º 37º 37º 37º 10 20 30 40 x(cm) 0 Sol. Here, we can say that the electric will be perpendicular to equipotential surfaces. r V |E| = Also, d Where, V = potential difference between two equipotential surfaces. d = perpendicular distance between two equipotential surfaces. r 60 |E| = So (10 sin 37º ) 10 2 = 1000 V/m Now there are two perpendicular directions: either direction 1 or direction 2 as shown in figure, but since we know that in the direction of electric field, electric potential decreases, so the correct direction is direction 2. Hence E = 1000 V/m, making an angle 127° with the x-axis Direction 2 Direction 1 Figure shows some equipotential surfaces produce by some charges. At which point, the value of electric field is greatest? (50 V) (40 V) (30 V) (20 V) Ex. B C A ELECTROSTATICS Sol. E is larger where equipotential surfaces are closer. ELOF are  to equipotential surfaces. In the figure, we can see that for point B, they are closer so E at point B is maximum. ELECTROSTATIC POTENTIAL ENERGY Electrostatic potential energy of a point charge due to many charges The electrostatic potential energy of a point charge at a point in electric field is the work done in taking the charge from reference point (generally at infinity) to that point without change in kinetic energy Its Mathematical formula is Source charges Test charges U = W P)ext|K = 0 = qV = – W P)ele Here, q is the charge whose potential energy is being calculated and V is the potential at its position due to the source charges. Note : Always put q and V with sign. Properties Ex. (i) Electric potential energy is a scalar quantity but may be positive, negative or zero. (ii) Its unit is same as unit of work or energy i.e joule (in S.. system). Some times energy is also given in electron-volts. Where, 1eV = 1.6 × 10–19 J (iii) Electric potential energy depends on reference point. (Generally Potential Energy at r=  is taken zero) The four identical charges (q each) are placed at the corners of a square of side a. Find the potential energy of one of the charges due to the remaining charges. Sol. The electric potential of point A due to the charges placed at B, C and D is 1 1 q V = 4 + 4 a 0 0  1 1  1  q q + 4 = 4  2   a 2 a 2a 0 0  q Potential energy of the charge at A is = qV = a q D q C a q a A B q 1  1  q2 2 .   40  2 a Ex. A particle of mass 40 mg and carrying a charge 5 × 10 –9 C is moving directly towards a fixed positive point charge of magnitude 10 –8 C. When it is at a distance of 10 cm from the fixed point charge it has speed of 50 cm/ s. At what distance from the fixed point charge will the particle come momentarily to rest? Is the acceleration constant during the motion? Sol. If the particle comes to rest momentarily at a distance r from the fixed charge, then from conservation of energy, we have ? Qq 1 Qq 1 1 mu 2 + = a r 4 4 2 0 0 Substituting the given data, we get: 1 1 1 × 40 × 10–6 × × = 9 × 109 × 5 × 10–8 ×10–9 2 2 2 1   r  10    5 106 100 1 1 190  –10 =  = 8 9 9  5 10 9 r r or, or, i.e., As here,  r= 9 m 190 r = 4.7 × 10–2 m F= 1 Qq 40 r 2 So, acc. = F 1  2 m r i.e., Acceleration is not constant during the motion. Ex. A proton moves from a large distance with a speed u m/s directly towards a free proton originally at rest. Find the distance of closest approach for the two protons in terms of mass of proton m and its charge e. Sol. As here the particle at rest is free to move, when one particle approaches the other, due to electrostatic repulsion other will also start moving and so the velocity of first particle will decrease while of other will increase and at closest approach, both will move with same velocity. So, if v is the common velocity of each particle at closest approach, then by 'conservation of momentum' of the two proton system. mu = mv + mv i.e., v= 1 u 2 And by conservation of energy, 1 1 1 1 mu2 = mv2 + mv2 + 2 2 2 4 0 e2 r 2  e2 1 1 u mu2 – m   = 2 4 0 r 2  r= [as v = u ] 2 2  e 1 mu2 = 4  4 0r e2 m 0 u 2 Electrostatic potential energy of a system of charges (This concept is useful when more than one charges move.) It is the work done by an external agent against the internal electric field required to make a system of charges in a particular configuration from infinite separation without change in their kinetic energies. Types of system of charges (i) (ii) Point charge system Continuous charge system. Derivation for a system of point charges (i) Keep all the charges at infinity. Now bring the charges one by one to its corresponding position and find work required. PE of the system is algebraic sum of all the works. Let W 1 = Work done in bringing first charge. W 2 = Work done in bringing second charge against force due to 1 st charge. W 3 = Work done in bringing third charge against force due to 1st and 2 nd charge. PE = W1 + W 2 + W 3 + ...... . (This will contain n(n  1) n = C2 terms) 2 ELECTROSTATICS (ii) Method of calculation (to be used in problems) : U = sum of the interaction energies of the charges. = (U12 + U13 + ........ + U1n) + (U23 + U24 + ........ + U2n) + (U34 + U35 + ........ + U3n) .... . (iii) Method of calculation useful for symmetrical point charge systems. Find PE of each charge due to rest of the charges. If U1 = PE of first charge due to all other charges. = (U 12 + U13 + ........ + U1n) U2 = PE of second charge due to all other charges. = (U21 + U23 + ........ + U2n) then U = PE of the system = U1  U2  ....Un 2 Ex. Find out potential energy of the two point charge system having charges q 1 and q 2 separated by distance r. Sol. Let both the charges be placed at a very large separation initially. Let, W 1 = work done in bringing charge q 1 in absence of q 2 = q 1(Vf – Vi) = 0 W 2 = work done in bringing charge q2 in presence of q 1 = q 2(Vf – V i) = q 2(Kq 1/r – 0)  Ex. PE = W 1 + W 2 = 0 + Kq 1q 2 / r = Kq 1q 2 / r Figure shows an arrangement of three point charges. The total potential energy of this arrangement is zero. Calculate the ratio Sol. Usys = 1  qQ (q)(q) Q(q)    =0 40  r 2r r  –Q + q –Q=0 2 or 2Q = -Q +q r q . Q q or 2 +q 2r q 4 = . Q 1 Ex. Two point charges, each of mass m and charge q are released when they are at a distance r from each other. What is the speed of each charged particle when they are at a distance 2r ? Sol. According to momentum conservation, both the charge particles will move with same speed. Now applying energy conservation :  1 Kq 2 Kq 2 0+0+ = 2 mv2 + 2 r 2r Ex.  v= V Kq 2 2rm V 2r Two charged particles each having equal charges 2 × 10 –5 C are brought from infinity to within a separation of10 cm. Calculate the increase in potential energy during the process and the work required for this purpose. Sol. U = Uf – Ui = Uf – 0 = Uf We have to simply calculate the electrostatic potential energy of the given system of charges q 1q 2 9  109  2 105  2 105  100 1 = J = 36 J r 10 40  U = Uf =  Work required = 36 J = equal to change in potential energy of system Ex. Three equal charges q each are placed at the corners of an equilateral triangle of side a. A q (i) Find out potential energy of charge system. (ii) Calculate work required to decrease the side of triangle to a/2. B q (iii) If the charges are released from the shown position and each of them has same mass m then find the speed of each particle when they lie on triangle of side 2a. Sol. (i) Method I (Derivation) qC a A Assume all the charges are at infinity initially. Work done in putting charge q at corner A  B W1 = q (vf – vi) = q (0 – 0) a C Since potential at A is zero in absence of charges, work done in putting q at corner B in presence of charge at A :   Kq  Kq 2  0 q = W2 =  a  a  Similarly work done in putting charge q at corner C in presence of charge at A and B.  W3 = q(vf – vi)  Kq Kq    = q  0 a    a = 2 Kq 2 a So, net potential energy PE = W1 + W2 + W3 =0+ Kq 2 2Kq 2 + a a = 3Kq 2 a Method II (using direct formula): U = U12 + U13 + U23 = Kq 2 Kq 2 Kq 2 3Kq 2 + + = a a a a (ii) Work required to decrease the sides W = U f – Ui = 3Kq 2 3Kq 2 3Kq 2 – = Joules a/2 a a (iii) Work done by electrostatic forces = Change is kinetic energy of particles. Ui – U f = Kf – Ki  Ex. 3Kq 2 3Kq 2 1 – = 3( mv2) – 0  v = a 2a 2 Kq 2 am Four identical point charges q each are placed at four corners of a square of side a. q q 1q q 4 3 Find out potential energy of the charge system Sol. Method 1 (using direct formula) : U = U12 + U13 + U14 + U23 + U24 + U34 = 2 Kq 2 Kq 2 Kq Kq 2 Kq 2 Kq 2 + + + + + a a a a a 2 a 2  4Kq 2 2Kq 2   =  a  a 2   = 2Kq 2 a 1   2   2  2 ELECTROSTATICS 1 (U + U2 + ......)] : 2 1 U1 = total P.E. of charge at corner 1 due to all other charges. Method 2 [Using, U = U2 = total P.E. of charge at corner 2 due to all other charges. U3 = total P.E. of charge at corner 3 due to all other charges. U4 = total P.E. of charge at corner 4 due to all other charges. Since, due to symmetry, U1 = U2 = U3 = U4  Kq 2 Kq 2 Kq 2  2Kq 2  1  U1  U 2  U 3  U 4  = UNet = = 2U1 = 2  a  a  2   2a  2 a 2   Ex. Sol. q Six equal point charges q each are placed at six corners of a hexagon of side a. Find out potential energy of charge system UNet = q q U1  U 2  U 3  U 4  U 5  U 6 2 q Due to symmetry U1 = U2 = U3 = U4 = U5 = U6 So Unet = 3U1 = 3Kq 2 a q q a  2 1   2  3 2  Derivation of electric potential energy for continuous charge system This energy is also known as self energy. (i) Finding P.E., (Self Energy) of a uniformly Charged spherical shell For this, lets use method 1 : Take an uncharged shell. Now bring charges one by one from infinity to the surface fo the shell. The work required in this process will be stored as potential Energy. R Charges uncharged shell Suppose, we have given charge q to the sphere and now we are giving extra charge dq to it. Work required to bring dq charge from infinity to the shell is  + + + +  q o Kq KQ 2 dq = R 2R This work will be stored in the form of P.E. (self energy) So, P.E. of a charged spherical shell: KQ 2 2R + + q Q U= q R dW = (dq) (Vf - Vi) Kq Kq dW = (dq) ( - 0) = dq R R  Total work required to give charge Q is W = + + + + dq (ii) Self energy of uniformly charged solid sphere In this case we have to assemble a solid charged sphere. So we bring the charges one-by-one from infinity to the sphere so that the size of the sphere increases. Suppose we have given charge q to the sphere, and its radius becomes ‘x’ .Now we are giving extra charge dq to it, which will increase its radius by ‘dx’  Work required to bring charge dq from infinity to the sphere  Kq   o  = Kqdq = dq (Vf - Vi) = (dq)   x  x  Total work required to give charge Q to the sphere W= &  Kqdq , x where dq =  (4 x dx)  2 4 3 q =   x  3  4    x 3   (4x 2 dx) 3   W =0 K x o x xR Solving, well get: 3 KQ 2 W= = Uself (for a solid sphere) 5 R Ex. A spherical shell of radius R with uniform charge q is expanded to a radius 2R. Find the work performed by the electric forces and external agent against electric forces in this process. Sol. W ext = Uf – Ui = q2 q2 q2 – =– 16  0 R 8  0 R 16  0 R q2 q2 q2 W elec = U i – Uf = – = 8  0 R 16  0 R 16  0 R Ex. Two non-conducting hollow uniformly charged spheres of radii Q1 R1 R1 and R2 with charge Q1 and Q2 respectively are placed at a distance r. Q2 R2 r Find out total energy of the system. Q1Q 2 Q12 Q 22 + + 4  r 8 0 R 1 8 0 R 2 0 Sol. Utotal = Uself + UInteraction = Ex. Two concentric spherical shells of radius R1 and R2 (R2 > R1) are having uniformly distributed charges Q1 and Q2 respectively. Find out total energy of the system. Sol. Utotal = Uself 1 + Uself 2 + UInteraction = Q12 80 R1 + Q 22 80 R 2 Q1Q2 + 40 R 2 R2 Q1 R1 Q2 ELECTROSTATICS Energy density Def : Energy density is defined as energy stored in unit volume in any electric field. Its mathematical formula is given as following : 1 2 E 2 E = electric field intensity at that point  = 0r electric permittivity of medium Energy density = where . Ex. Sol. Find out energy stored in an imaginary cubical volume of side a in front of a infinitely large non-conducting sheet of uniform charge density . Energy stored : U=    Ex. Sol. 1  2 0 E 2 dV ; where dV is small volume 1 0 E 2  dV 2 E is constant . U= 1 2 2 a 3 U = 2 0 2 . a3 .= 80 4 0 Find out energy stored in the electric field of uniformly charged thin spherical shell of total charge Q and radius R. We know that electric field inside the shell is zero, so the energy is stored only outside the shell, which can be calculated by using energy density formula. x  1 0 E 2dV 2 Consider an elementary shell of thickness dx and radius x (x > R). Volume of the shell = (4x2dx) = dV  x R Uself = U Ex. Sol.  2 1  KQ  1 0 . 4x2 dx = 0 K 2 Q 2 4 2  x 2  2 = R = KQ2 Q2 Q2 40 1 . = = .   2R 80 R 2 (40 )2  R  x dx 1 dx x2  R Find out energy stored inside a solid non-conducting sphere of total charge Q and radius R. [Assume charge is uniformly distributed in its volume.] We can consider solid sphere to be made of large number of concentric spherical shells. Also electric field intensity at the location of any particular shell is constant. R1 2  Uinside =  0 0 E dV 2 x Consider an elementary shell of thickness dx and radius x. Volume of the shell = (4x2dx) Uinside = = R 0 2 1  KQx  1 K 2Q 2 4 0  3  . 4x2 dx = 0 2 2  R  R6 40 2R 6 Q2 Q2 KQ2 R5 . = = . 400 R 10 R (40 )2 5 R 0 x 4dx dx Relation between Electric Field Intensity and electric Potential (i) For uniform electric field Potential difference between two points A and B B  VB – VA = – E . AB (ii) A Non uniform electric field Ex = – V V V , Ey = – , Ez = – y z x r E = Ex î + Ey ĵ + Ez k̂  ˆ  ˆ  ˆ  ˆ   ˆ  ˆ   = – i V  j V  k V  = – i  j  k  V y z  y z   x  x = – V = –grad V V = derivative of V with respect to x (keeping y and z constant) x Where, V = derivative of V with respect to y (keeping z and x constant) y V = derivative of V with respect to z (keeping x and y constant) z If electric potential and electric field depends only on one coordinate, say r (i) (ii) r V r̂ E =– r where, r̂ is a unit vector along increasing r.. r uur  dV = –  E  dr rB  VB – VA = – r   E. dr rA uur dr is along the increasing direction of r.. (iii) The potential of a point r V=– r   E. dr  Ex. A uniform electric field is along x – axis . The potential difference VA– VB = 10 V is between two points A (2m , 3m) and B (4m, 8m). Find the electric field intensity. Sol. E= Ex. V = x2 + y , Find E . Sol. V = 2x, x V 10 = = 5 V / m. d 2 (It is along + ve x-axis) r V 1 y and V 0 z  ˆ V ˆ V ˆ V  r E = –  i x  j y  k z  = –(2x î + ĵ )    Electric field is non-uniform. ELECTROSTATICS Ex. Sol. r For given E = 2xiˆ  3yjˆ , find the potential at (x, y) if V at origin is 5 volts. v y x r dV   E  dr E dx = – –    E ydy  x 5  0 0 V–5=– 2x 2 3y 2  2 2 V=–  2x 2 3y 2  + 5. 2 2 Electric Dipole If two point charges, equal in magnitude ‘q’ and opposite in sign separated by a distance ‘a’ such that the distance of field point r>>a, the system is called a dipole. The electric dipole moment is defined as a vector quantity having magnitude p = (q × a) and direction from negative charge to positive charge. ETOOS KEY POINTS [In chemistry, the direction of dipole moment is assumed to be from positive to negative charge.] The C.G.S unit of electric dipole moment is debye which is defined as the dipole moment of two equal and opposite point charges each having charge 10 –10 Franklin and separation of 1 Å, i.e., 1 debye (D) = 10–10 × 10–8 = 10 –18 Fr × cm or 1 D = 10–18 × C × 10–2 m = 3.3 × 10–30 C × m. 3 109 S.I. Unit is coulomb × metre = C . m Ex. A system ha s t wo charges q A = 2. 5 × 1 0 –7 C and q B = – 2. 5 × 1 0 –7 C lo cat ed at p oints A : (0, 0, – 0.15 m) and B ; (0, 0, + 0.15 m) respectively. What is the net charge and electric dipole moment of the system ? Sol. Net charge = 2.5 × 10–7 – 2.5 × 10 –7 = 0 Electric dipole moment, P = (Magnitude of charge) × (Separation between charges) = 2.5 × 10–7 [0.15 + 0.15] C m = 7.5 × 10–8 C m The direction of dipole moment is from B to A. ELECTRIC FIELD INTENSITY DUE TO DIPOLE (i) At the axial point :- a -q +q r P r E Kq 2  a r   2   If r >> a then, Kq a  r   2  2 r r 2KP Kq 2ra P̂ = , E= 4 r3 r ur (along the P) = Kq (2ra)  2 a2 r   4      2 P̂  As the direction of electric field at axial position is along the dipole moment ( P ) So, (ii)   E axial 2K P = r3 Electric field at perpendicular Bisector (Equatorial Position) Enet = 2 E cos  (along – P̂ )  Enet     a   Kq Kqa   2 ˆ = ˆ (P) (P) = 2 2 2 2 3/2 2   a   r2   a  2 a  r2         r      2    2    2        If r >> a then  KP ˆ Enet = 3 (P) r   As the direction of E at equatorial position is opposite of P so we can write in vector form:   KP r3 Electric field at general point (r, ) : E eqt = – (iii)  For this, let’s resolve the dipole moment P into components. One component is along radial line (=P cos) and other component is r to the radial line (=Psin) ELECTROSTATICS 2  2KP cos    KPsin   Er 2  E t 2      r3    r3  From the given figure Enet = Ex. Sol. Ex. Sol. Sol. = 1 2p 1 p = 3 40 r 40 r '3 2 1 3 = r r´3 or or r3 =2 r´3 or, r = 21/3 r´ Two charges, each of 5 C but opposite in sign, are placed 4 cm apart. Calculate the electric field intensity of a point that is at a distance 4 cm from the mid point on the axial line of the dipole. We cannot use formula of short dipole here because distance of the point is comparable to the distance between the two point charges. q = 5 × 10–6 C, a = 4 ×10–2 m, r = 4 × 10–2 m K(5C)  2cm  2 – K(5C)  6 cm  2 = 144 NC–1 = 108 N C–1 144  10 8 Two charges ± 10 C are placed 5 × 10 –3 m apart as shown in figure. Determine the electric field at a point Q which is 0.15 m away from O, on the equatorial line. In the given problem, r >> a 1 q(a) 1 p = 40 r 3 40 r 3  E= or E = 9 × 109 × 10  106  5  103 NC–1 0.15  0.15  0.15 = 1.33 ×105 NC–1 ELECTRIC POTENTIAL DUE TO A SMALL DIPOLE (i) KP 1  3cos 2  r3 KPsin  tan  Et r3  tan  = = 2KP cos  2 Er r3 KP tan  2 Enet = 3 1  3cos  ; tan  = 2 r The electric field due to a short dipole at a distance r, on the axial line, from its mid point is the same as that r of electric field at a distance r', on the equatorial line, from its mid-point. Determine the ratio . r´ Eres = E + + E – = Ex. 2 Potential at axial position : V= V= Kq K(q)  a    a r   r    2  2 Kqa  2  a 2  r      2    If r >> a then V= Kqa ; r2 where, qa = p  Vaxial = KP r2 (ii) Potential at equatorial position : V= Kq a r   2 2 2 K(q)  a r   2 2 =0 2 Veqt = 0 (iii) Potential at general point (r,) :  Lets resolve the dipole moment P into components: Pcos along radial line and Psin r to the radial line For the Pcos component, the point A is an axial point, So, potential at A due to Pcos = K  P cos   r2 And for Psin component, the point A is an equatorial point, So potential at A due to Psin = 0 Vnet =  K  P cos   r2   K  P . r    V= r3 Ex. (i) Find potential at point A and B due to the small charge - system fixed near origin.(Distance between the charges is negligible). (ii) Find work done to bring a test charge q0 from point A to point B, slowly. All parameters are in S.I. units. Sol. (i) Dipole moment of the system is r P = (qa) î + (qa) ĵ Potential at point A due to the dipole r r ˆ (P· r) K[(qa) ˆi  (qa) ˆj]·(4iˆ  3j) k(qa) 7 VA = K 3 = = 3 125 r 5  K  (qa) ˆi  (qa) ˆj · 3 ˆi – 4 ˆj  VB = (ii)  K(qa)  K(qa) (7)   –K q q 0 a – WA B = UB – UA = q0 (VB – VA) = q0  – (8)    WA B =  125    125 125 (5) 3 = –K (qa) 125  ELECTROSTATICS DIPOLE IN UNIFORM ELECTRIC FIELD (i) Dipole is placed along electric field : –q qE +q qE E In this case, Fnet = 0, net = 0, so it is an equilibrium state. And it is a stable equilibrium position. (ii)  If thedipoleisplaced at angle from E : In this case Fnet = 0 but Net torque  = (qEsin) (a) Here qa = P   = PE sin    in vector form :   P  E Ex. Sol. A dipole is formed by two point charge –q and +q, each of mass m, and both the point charges are connected by a  rod of length  and mass m. This dipole is placed in uniform electric field E . If the dipole is disturbed by a small angle  from stable equilibrium position, prove that its motion will be almost SHM. Also find its time period. If the dipole is disturbed by  angle, net = –PE sin (Here – ve sign indicates that direction of torque is opposite to ) If  is very small, sin    net = –(PE) net  (–) so motion will be almost SHM & C = PE (where, P = q)  T = 2  C 2  (iii) T = 2m– ml l   2m   12  2  = 2 P.E ml ml  12 2 ql E 2 =2 Potential energy of a dipole placed in uniform electric field : Br  UB – UA = –  F . dr (for translational motion) A Here, B  r UB – UA = –   .d (for rotational motion) A In the case of dipole, at  = 90° , P.E. is assumed to be zero. 7m l 2 = 2 12q l E 7m l 12qE = T= 7m l 3qE  U – U90° = –  (PEsin )(d) (As the direction of torque is opposite of ) 90 U – 0 = – PE cos   = 90° is chosen as reference, so that the lower limit comes out to be zero.   U = – P  E From the potential energy curve, we can conclude : Ex. (i) At  = 0, there is minimum of P.E. so it is a stable equilibrium position. (ii) At = 180° , there is maxima of P.E. so it is a position of unstable equilibrium. Two point masses of mass m and equal and opposite charge of magnitude q are attached on the corners of a nonconducting uniform rod of mass m and the system is released from rest in uniform electric field E as shown in figure from  = 53° (i) (ii) Sol. (iii) Find angular acceleration of the rod just after releasing What will be angular velocity of the rod when it passes through stable equilibrium. Find work required to rotate the system by 180°. (i) net = PE sin53° = I   4 (ql ) E   48qE 5  = 2 2 2 35 ml ml l  l   m   m  12 2 2 (ii) From energy conservation : Ki + Ui = Kf + Uf  0 + (– PE cos 53°) = 1 2  + (–PE cos 0°) 2 2 where I = (iii) 2 l  l  ml 2 7ml 2 +m  + m  = 12 12 2 2   1 7m l 2 2 × ×2 = qE or 2 5 12  Wext = Uf – Ui  Wext = (–PE cos(180° + 53°)) – (–PEcos 53°) or  3  3 Wext = (q)E   + (q)E   5 5 =  1 2 2  = PE (1– 3/5) = PE 2 5 48qE 35 ml 6 Wext =   qE 5 ELECTROSTATICS Dipole in Non-uniform Electric Field  If the dipole is placed along E , (shown in figure) Then, Net force on the dipole : Fnet = q E(x + dx) – q E(x) Fnet = q E (x  dx) – E(x) (dx) dx ; here (q (dx) = P)  dE    Fnet = P   dx  Ex. A short dipole is placed on the axis of a uniformly charged ring (total charge –Q, radius R) at a distance centre of ring as shown in figure. Find the Force on the dipole due to the ring Sol.   dE  F=P    dx   F=P  KQx d   2 2 3/2  ; dx  (R  x )  Solving we get, (at x = R from 2 R ) 2 F=0 Force between a dipole and a point charge Ex. A short dipole of dipole moment P is placed near a point charge Q as shown in figure. Find force on the dipole due to the point charge Sol. +Q F F –q P +q Force on the point charge due to the dipole F = (Q) Edipole  2KP  F = (Q)  3  (towards right)  r  From action reaction concept, force on the dipole due to point charge will be equal to the force on charge due to dipole F= 2KPQ (towards left) r3 is force on dipole due to point Charge. FORCE BETWEEN TWO DIPOLES Ex. A short dipole of dipole moment P is placed near a point charge Q as shown in figure. Find force on the dipole due to the point charge. +q +Q r -q Sol. Force on the point charge due to dipole F = (Q) (Edipole)  KP  F = (Q)  3    r   So force on the dipole due to the point charge will also be  KPQ  F =  3   (but in opposite direction) as shown  r   Ex. Find force on short dipole P2 due to short dipole P1 if they are placed at a distance r a part as shown in figure. r P2 P1 Sol. Force on P2 due to P1  dE1  F2 = (P2)    dr    d  2 kP1   F2 = (P2)   3    dr  r   6 K P1P2 r4 Here – sign indicates that this force will be attractive (opposite to r) or Ex. F2 = – Find force on short dipole P2 due to short dipole P1 if they are placed a distance r apart as shown in figure. r P1 Sol. P2 ELECTROSTATICS Fnet = q2 E (x + dx) - q2 E (x)  E(x  dx)  E(x)  Fnet = q2   dx dx    dE  Fnet = (P2)    dx  or  (Usually this formula is valid when the dipole is placed along E . However, in this case also, we are getting the same formula)  d  KP1    Fnet = (P2)   3   dr  r    Fnet = 3KP1P2 (in magnitude) & (direction upwards) r4 Electric Lines of Force (ELOF) The line of force in an electric field is an imaginary line, the tangent to which at any point on it represents the direction of electric field at the given point. Properties (i) Line of force originates out from a positive charge and terminates on a negative charge. If there is only one positive charge, then lines starts from positive charge and terminates at . If there is only one negative charge, then lines starts from  and terminates at negative charge. (ii) + – ELOF of Isolated positive charge ELOF of Isolated negative charge  Two lines of force never intersect each other because there cannot be two directions of E at a single Point impossilble (iii) Electric lines of force produced by static charges do not form closed loop. If lines of force make a closed loop, then work done to move a +q charge along the loop will be non-zero. So it will not be conservative field. So these type of lines of force are not possible in electrostatics. (iv) (v) Ex. Sol. (vi) (vii) Ex. Sol. Ex. Sol. The Number of lines per unit area (line density) represents the magnitude of electric field. If lines are dense  E will be more If Lines are rare  E will be less and if E = O, no line of force will be found there Number of lines originating (terminating) at a charge is proportional to the magnitude of charge If number of electric lines of force from charge q are 10, then find out number of electric lines of force from 2q charge. No. of ELOF  charge 10  q  20  2q So, number of ELOF will be 20. Electric lines of force end or start perpendicularly on the surface of a conductor. Electric lines of force never enter into conductors. Some electric lines of force are shown in figure. For points A and B (A) EA > EB (B) EB > EA (C) VA > VB (D) VB > VA Lines are more dense at A, so EA > EB In the direction of Electric field, potential decreases so VA > VB A B If a charge is released in electric field, will it follow lines of force? Case I If lines of force are parallel (in uniform electric field) :F = q0E +q0 E  In this type of field, if a charge is released, force on it will be qoE and its direction will be along E .So the charge will move in a straight line , along the lines of force. Case II If lines of force are curved (in non-uniform electric field) :+q The charge will not follow lines of force Ex. Sol. A charge + Q is fixed at a distance d in front of an infinite metal plate. Draw the lines of force indicating the directions clearly. There will be induced charge on two surfaces of conducting plate, so ELOF will start from +Q charge and terminate at conductor and then will again start from other surface of conductor. ELECTROSTATICS Electric Flux r dS Consider some surface in an electric field E . Let us select a small area element uur dS on this surface. The electric flux of the field over the r uur area element is given by dE = E.dS uur Direction of dS is normal to the surface. It is along n̂ or dE = EdS cos  or dE = (E cos ) dS or dE = En dS uur where E n is the component of electric field in the direction of dS . r uur The electric flux over the whole area is given by E =  E.dS =  E n dS S r r If the electric field is uniform over that area then E = E  S S Special Cases Case I E S If the electric field is normal to the surface, E  then angle of electric field E with normal will be zero So  = ES cos 0 or  = ES Case II If electric field is parallel of the surface (grazing),  then angle made by E with normal = 90º So  = ES cos 90º = 0 S Physical Meaning E The electric flux through a surface inside an electric field represents the total number of electric lines of force crossing the surface. It is a property of electric field Unit (i) (ii) Ex. Sol. Ex. Sol. The SI unit of electric flux is Nm2 C–1 (Gauss) or J m C–1. Electric flux is a scalar quantity. (It can be positive, negative or zero) r The electric field in a region is given by E  3 E 0ˆi  4 E 0ˆj with E0 = 2.0 × 103 N/C. Find the flux of this field through 5 5 a rectangular surface of area 0.2m2 parallel to the Y–Z plane. r r  3 ˆ 4 ˆ N  m2 E = E  S =  E 0 i  E0 j  . 0.2iˆ = 240 5 5  C   A point charge Q is placed at the corner of a square of side a, then find the flux through the square. The electric field due to Q at any point of the square will be along the plane of square and the electric field lines are perpendicular to square ; so  = 0. In other words we can say that no line is crossing the square so, flux = 0. Q a a Ex. Sol. Find the electric flux due to a point charge 'Q' through the circular region of radius R if the charge is placed on the axis of ring at a distance x. x We can divide the circular region into small rings. Lets take a ring of radius r and width dr. Flux through this small element d = E ds cos  net =  Case-III R +Q  r R E ds cos  =  r o KQ (2r dr) (x  r 2 ) 2  x  2 2  x r  Q  = 2 0    x 1  2  x  R 2   Curved surface in uniform electric field Suppose a circular surface of radius R is placed in a uniform electric field as shown. E Flux passing through the surface  = E (R ) 2 (ii) R Now suppose, a hemispherical surface, is placed in the electric field. Flux through hemispherical surface:  =  Eds cos   = E  ds cos  where,  ds cos  is projection of the spherical surface Area on base. &  ds E E R ds cos = R2 So,  = E(R2) = same Ans. as in previous case So, we can conclude that If the number of electric field lines passing through two surfaces are same, then flux passing through these surfaces will also be same, irrespective of the shape of surface = 1 Case IV = 3 = E(R2) 2 Flux through a closed surface Suppose there is a spherical surface and a charge 'q' is placed at centre.  Flux through the spherical surface   =  = 1 Q 4 R 2  E . ds =     E ds (as E is along ds (normal)) ds  1 Q  2 =  2   4 R   4 0 R  where,   = ds = 4 R 2 Q  ELECTROSTATICS Now if the charge Q is enclosed by any other closed surface, still same no of lines of force will pass through the surface. So, here also flux will be  = Q . That's what Gauss Theorem is.  Q =   Q =   +Q +Q Gauss's Law in Electrostatics or Gauss's theorem This law was stated by a mathematician Karl F Gauss. This law gives the relation between the electric field at a point on a closed surface and the net charge enclosed by that surface. This surface is called Gaussian surface. It is a closed hypothetical surface. Its validity is shown by experiments. It is used to determine the electric field due to some symmetric charge distributions. Statement and Details Gauss's law is stated as given below : The surface integral of the electric field intensity over any closed hypothetical surface (called Gaussian 1 times the total charge enclosed within the surface. Here,  0 is the 0 surface) in free space is equal to permittivity of free space. If S is the Gaussian surface and to Gauss's law, E = n q i i 1 is the total charge enclosed by the Gaussian surface, then according 1 r uur = E  dS Ñ  0 n q i i 1 (i) Flux through Gaussian surface is independent of its shape. (ii) Flux through Gaussian surface depends only on total charge present inside Gaussian surface. (iii) Flux through Gaussian surface is independent of position of charges inside Gaussian surface. (iv) Electric field intensity at the Gaussian surface is due to all the charges present inside as well as outside the Gaussian surface. (v) In a closed surface incoming flux is taken negative, while outgoing flux is taken positive, because n̂ is (vi) In a Gaussian surface,  = 0 does not imply E = 0 at every point of the surface but E = 0 at every point taken positive in outward direction. implies  = 0. Ex. Find out flux through the given Gaussian surface. Sol. = Ex. If a point charge q is placed at the centre of a cube, then find out flux through any one face of cube. Sol. Flux through all 6 faces = 2C  3C  4C 3  106 Qin = = Nm2/C 0 0 0 q . Since, all the surfaces are symmetrical 0 So, flux through one face = 1 q 6 0 Flux through open surfaces using Gauss’s Theorem Ex. A point charge +q is placed at the centre of curvature of a hemisphere. Sol. +q Find flux through the hemispherical surface. Lets put an upper half hemisphere. Now, flux passing through the entire sphere = q 0 As the charge q is symmetrical to the upper half and lower half hemispheres , so half-half flux will emit from both the surfaces. Flux emitting q from lower half surface = 2 0 Flux emitting q from upper half surface = 2 Ex. A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge a as shown in figure. Find the flux of the electric field through the square surface. Sol. We can consider imaginary faces of cube such that the charge lies at the centre of the cube. Due to symmetry, we can say that flux through the Q given area (which is one face of cube),  = 6 0 Ex. Find flux through the hemispherical surface Sol. (i) Flux through the hemispherical surface due to +q = • 0 +q a/2 a a Q +q q (we have seen in previous examples) 20 (ii) Flux through the hemispherical surface due to +q0 is 0, because due to +q0, field lines entering the surface = field lines coming out of the surface. +q0 ELECTROSTATICS Finding qin from flux Ex. Flux (in S.I.units) coming out and entering a closed surface is shown in the fig- 30 ure . Find charge enclosed by the closed surface. Sol. Net flux through the closed surface = + 20 + 30 + 10 -15 = 45 N.m2/c From Gauss`s theorem : q in net =  0  or q in 45 =  0 15 qin = (45) Finding electric field from Gauss`s Theorem From Gauss`s theorem, we can say q in =  = E.ds net 0    Finding E due to a spherical shell (a) Electric field outside the Sphere Since, electric field due to a shell will be radially outwards. So lets choose a spherical Gaussian surface Applying Gauss`s theorem for this spherical Gaussian surface, q in q =  = = E ds net           | E || ds | cos 0 because  E is normal to the surface  E  ds (because value of E is constant at the surface) E (4r2) (  ds total area of the spherical surface = 4 r2) q in E (4r2) =  0  (b)  q Eout = 4 r 2  Electric field inside a spherical shell Lets choose a spherical Gaussian surface inside the shell. Applying Gauss`s theorem for this surface   E ds =  net q in =  =0 0     | E || ds | cos 0  E  ds   E (4r2) E (4r2) = 0  20 10 Ein = 0 Electric field due to solid sphere (having uniformly distributed charge Q and radius R) (a) Electric field outside the sphere Direction of electric field is radially outwards, so we will choose a spherical Gaussian surface Applying Gauss`s theorem q in Q  =  = = E ds net      Q,R + +   | E || ds | cos 0  E  ds  E (4r2)  (b) Q E (4r2) =    or Q Eout = 4 r 2  Electric field inside a solid sphere : ++ + ++ + + + + + + + + + + + + ++ + + ++ + + + ++ + + Q,R + + + + r+ ++ ++ + ++ + ++ + + + + + + + ++ ++ ++ Q = Total charges contained by solid sphere R = Radins of sphere G.S For this choose a spherical Gaussian surface inside the solid sphere Applying Gauss`s theorem for this surface q in =  = E ds net    Q 4  r 3 4 3 3 R Qr 3 3 = =   R 3   E ds  E (4r2)  E(4r2) = Q r E = 4 R 3    Ein = Qr 3  R 3 kQ r R3 + + + + + + + + + ELECTROSTATICS Electric field due to infinite line charge (having uniformly distributed charged of charge density l) Electric field due to infinitely wire is radial so we will choose cylindrical Gaussian surface as shown is figure: q in l =  =    3 =    E.ds =  E ds = E  ds = E (2r) l  E (2r) =      2k E = 2 r = r  Electric field due to infinitely long charged tube (having uniform surface charge density s and radius R)) (i) E outside the tube :- Lets choose a cylindrical Gaussian surface of length  : q in 2Rl net =  =    (ii)  Eout × 2r = 2Rl  E inside the tube : Lets choose a cylindrical Gaussian surface inside the tube. q in net =  = 0  So Ein = 0  R E= r  0 E due to infinitely long solid cylinder of radius R (having uniformly distributed charge in volume (volume charge density )) : (i) E at outside point :Lets choose a cylindrical Gaussian surface. Applying Gauss`s theorem : q in  R 2l E × 2r =  =   Eout = (ii) R 2 2r 0 E at inside point : Lets choose a cylindrical Gaussian surface inside the solid cylinder. Applying Gauss`s theorem : E × 2r = q in   r 2 l =   r Ein = 2  0 Conductor and it's properties [For electrostatic condition] (i) (ii) (iii) (iv) (v) (vi) (vii) Conductors are materials which contain large number of free electrons which can move freely inside the conductor. n electrostatics conductors are always equipotential surfaces. Charge always resides on outer surface of conductor. f there is a cavity inside the conductor having no charge then charge will always reside only on outer surface of conductor. Electric field is always perpendicular to conducting surface. Electric lines of force never enter into conductors. Electric field intensity near the conducting surface is given by formula uur  n̂ E= 0 A C (viii) B uuur  uuur  uuur  E A  A nˆ ; E B  B nˆ and E C  C n̂ 0 0 0 When a conductor is grounded its potential becomes zero. V=0 ELECTROSTATICS (ix) (x) (xi) When an isolated conductor is grounded then its charge becomes zero. When two conductors are connected there will be charge flow till their potentials become equal. Electric pressure : Electric pressure at the surface of a conductor is given by formula P= 2 , where  is the local surface charge density.. 2 0 + + + + A + + Finding field due to a conductor + Suppose we have a conductor and at any 'A', local surface charge density =  We have to find electric field just outside the conductor surface. For this, let’s consider a small cylindrical Gaussian surface, which is partly inside and partly outside the conductor surface, as shown in figure. E=? + + + + + + + + + + + It has a small cross section area ds and negligible height. Applying Gauss's theorem for this surface : So, Eds = ds 0   E=  0 Electric field just outside the surface of conductor : E=  (direction will be normal to the surface) 0 r  in vector form: E  nˆ 0 (Here, n̂ = unit vector normal to the conductor surface) Electrostatic pressure at the surface of the conductor + uppose a conductor is given some charge. Due to repulsion, all the charges will reach the surface of the conductor. But the charges will still repel each other. So an outward force will be felt by each charge due to others. Due to this force, there will be some pressure at the surface, which is called electrostatic pressure. To find the electrostatic pressure, lets take a small surface element having Area 'ds'. Force on this element due to the remaining charges :  electric field at    that place due to  dF =   remainig charges     charge of   the small  element       + + + + + + + + + + + + + + + + + Let electric field at that point due to the remaining charges = Er and charge of the small element = dq = ds  dF = (Er) (dq) = (Er) (ds) So, pressure on this small element dF (E r ) (ds)  ds ds  P = (Er) () P ...........(1) Now to find pressure, we have to find Er (electric field at that position due to the remaining charges) Suppose, Electric field due to the small element near the surface = Es Electric field due to the remaining part near the surface = Er At a point just outside the surface, electric field due to the small element (Es) will be normally out wards, and electric field due to the remaining part (Er) will also be normally out wards. So Net electric field just outside the surface = Es + Er and we have proved that Es Er ds  electric field just outside the conductor surface =  o   Es + Er =  o ...........(2) ds Now, lets see the electric field just inside the metal surface. Here, electric field Es Er due to the remaining charges (Er) will be in the same direction (normally outward), but the electric field due to the small element will be in opposite direction (normally inward) So net electric field just inside the metal surface = Er – E s and we know that electric field inside a conductor = 0 So, E r – Es = 0  Er = Es ...........(3) from eqn. (2) and eqn. (3), we can say that :  2Er =  o   E r = 2 o Now, we can easily find the pressure from eqn. (1)  2 P = (Er)() = 2 () = 2 o o So, electrostatic pressure at the surface of the conductor P = where, = local surface charge density. 2 2 o ELECTROSTATICS 1. Electric Charge Charge of a material body is that property due to which it interacts with other charges. There are two kinds of charges-positive and negative. S.I. unit is coulomb. Charge is quantized, conserved, and additive. 2. Coulomb`s law r Force between two charges F  q1q 2 1 ˆr 4 0 r r 2 r = dielectric constant q1 q2 r Note : The Law is applicable only for static and point charges. Moving charges may result in magnetic interaction. And if charges are extended, induction may change the charge distribution. 3. Principle Of Superposition Force on a point charge due to many charges is given by r r r r F  F1  F2  F3  ..... Note : 4. The force due to one charge is not affected by the presence of other charges. Electric Field or Electric Intensity or Electric Field Strength (Vector Quantity) In the surrounding region of a charge there exist a physical property due to which other charge experiences a force. The direction of electric field is direction of force experienced by a positively charged particle and the magnitude of the field (electric field intensity) is the force experienced by a unit charge. r ur F E  unit is N/C or V/m. q 5. Electric field intensity due to charge Q 6. Null point for two charges r ur F 1 Q E  Lim  rˆ q0 0 q 4 0 r 2 0 r Q1 Q2  Null point near Q2 x Q1 r Q1  Q 2 (+) for like charges (–) for unlike charges If |Q1| > |Q2| 7. Equilibrium of three point charges r (i) Two charges must be of like nature. x 8. Q1 Q1  Q 2 r q r Q1 (ii) Third charge should be unlike nature. q and  Q1Q 2 Q1  Q 2  Q2 2 Equilibrium of symmetric geometrical point charged system q a a a Q q a a q q q Q q a a q Value of Q at centre for which system to be in state of equilibrium (a) For equilateral triangle Q  9. q (b) For square Q  3   q 2 2  1 4 Equilibrium of suspended point charge system For equilibrium position Tcos  = mg & Tsin  = Fe  Fe kQ 2 kQ 2  tan    x2 mg x 2 mg If whole set up is taken into an artifical satelite (geff ; 0) T = Fe = 10. Electric potential difference 11. Electric potential = W/q It is the work done against the field to take a unit positive charge from infinity (reference point) to the given point (i) For point charge : V = K 12. Relation between = – grad 13. (ii) For several point charges : V = K &V , ; ,V= Electrical potential energy of two charges : Add. 41-42A, Ashok Park Main, New Rohtak Road, New Delhi-110035 +91-9350679141 70 ELECTROSTATICS 14. 15. Electric dipole (a) Electric dipole moment p = qd (b) Torque on dipole placed in uniform electric field (c) Work done in rotating dipole placed in uniform electric field (d) Potential energy of dipole placed in an uniform field U = (e) At a point which is at a distance r from dipole midpoint and making angle  with dipole axis. (1) Potential (2) Electric field (3) Direction (f) Electric field at axial point (or End-on) (g) Electric field at equatorial position (Broad-on) of dipole of dipole Equipotential Surface And Equipotential Region In an electric filed the locus of points of equal potential is called an equipotential surface. An equipotential surface and the electric field line meet at right angles. The region where E = 0, Potential of the whole region must remains constant as no work is done in displacement of charge in it. It is called as equipotential region like conducting bodies. 16. Mutual Potential Energy Or Interaction Energy “The work to be done to integrate the charge system”. For 2 particle system For 3 particle system For n particle there will be terms. Total energy of a system = Uself + Umutual 17. Electric flux : (a) For uniform electric field ; = EA cos where  = angle between & area vector is contributed only due to the component of electric field which is perpendicular to the plane. (b) 18. If is not uniform throughout the area A, then Gauss`s Law : (Applicable only to closed surface) Net flux emerging out of a closed surface is . where qen = net charge enclosed by the closed surface (a) Shape and size of the closed surface (b) The charges located outside the closed surface. 19. For a conducting sphere (a) For 20. For a non - conducting sphere (a) For (b) For (b) For does not depend on the . Flux ELECTROSTATICS 21. For a conducting/non conducting spherical shell (i) For (ii) For 22. For a charged circular ring , Electric field will be maximum at 23. For a charge long conducting cylinder (a) For (b) For r < R : E = 0 24. Electric field intensity at a point near a charged conductor 25. Mechanical pressure on a charged conductor 26. For non-conducting long sheet of surface charge density 27. For conducting long sheet of surface charge density 28. Energy density in electric field 29. Electric lines of force Electric lines of electrostatic field following properties (a) Imaginary (b) Can never be closed each other (c) Can never be closed loops + A B – qA>qB (d) The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In rationalised MKS system (1/  0 ) electric lines are associated with it (called flux) will be q/  0 . (e) Lines of force ends or starts normally at the surface of a conductor. (f) If there is no electric field there will be no lines of force. (g) Lines of forceper unit area normal to the area at a point represents magnitude of intensity, crowded lines represent strong field while distant lines weak field. (h) Tangent to the line of force at a point is an electric field gives the direction of intensity.