Chapter 1: Electric Charges and Fields

Ch-1 Electric charges & Fields Electrostatics deals with study of forces, fields & potentials arising from static charges. Electric charge - It is something possessed by material object that makes it possible for them to exert electrical force & to respond to electrical force. Look at some experiment: i) A glass rod rubbed with a piece of silk brought close to suspended glass rod rubbed with silk repel the latter. So we can say, the glass rod acquire electric charge on rubbing. ii) Similar happen with ebonite rod. iii) But when glass & ebonite rod brought closer they attract. So we can say charge develop on glass rod rubbed with silk is different from ebonite rod’s charge. When glass rod rubbed with silk, rod acquire one kind of charge & silk acquire second kind of charge. This is true for any pair of object that are rubbed to electrified. Further, like charges repel each other, & unlike charges attract each other. Page 3 Charges acquired by object on rubbing against each other must be equal & opposite. By convention, charge acquired by glass rod is called positive, & by ebonite rod is negative. The property which differentiates two kinds of charges is called polarity of charge. Gold leaf Electroscope For detecting presence of electric charge & its polarity. Can also be used for measuring potential difference. It consists of 2 extremely thin gold foils (LL) attached to lower end of metal rod. When charged rod touched with metal disc D, the same charge is transferred to gold leaf through metal rod. The leaves repel each other. By measuring divergence of leaves, amount of charge can be estimated. Page 4 Origin of Electric charge in electrostatics Every object contains vast hidden charge, which is usually hidden as object contains equal amounts of positive & negative charges. Because of this equality or balance of charge, object is said to be electrically neutral, i.e. it contains no net charge. When we rub two insulating substance against each other, we provide energy to overcome friction b/w them. This energy is used in removing electrons from one substance & transferring them to other. Transfer take place from material in which electrons are held less tightly (with lower work function) to material in which electrons are held more tightly (with higher work function). i.e. cause of charging is actual transfer of electron. positively charged body has lost some electrons & hence its mass reduce slightly & vice versa. Page 5 Conductors, Insulators & Dielectrics Substance which can be used to carry or conduct electric charge from one place to other is called a conductor. e.g. Silver, gold, iron, etc. Earth & Human body is also a good conductor. In metallic conductors, there are very large numbers of free electrons, which act as carriers of charge. These free e- move are outer (valence) e- & move randomly in different direction. In an external electric field, free e- drift against direction of field, i.e. they get aligned. In electrolytic conductors, the charge carriers are both, +ive & -ive ions. Material which cannot conduct electricity, i.e. they are poor conductor of electricity, e.g. Rubber, plastic, wax, paper, etc. Such substance possess almost no free e-. Dielectrics are insulating materials which transmits electric effects without conducting. Dielectrics cannot conduct electricity. However induced charge appear on surface of dielectric when an external electric field is applied. Page 6 Charging by Induction When a charged body A imparts to another body B, some charge of opposite sign without any actual contact b/w A & B. Obviously, body A shall not lose any charge as it is not in contact with B. This is charging by Induction. Charging metal sphere by inductions: (Diagrams 1 to 5 showing the process of bringing a rod near a sphere, grounding it, and leaving it charged). In this process, positively glass rod does not lose any charge or electron. This is contrary to charging by conduction, i.e. charging by actual contact where the charged glass rod loses some charge. Quantization of Electric charge It is the property by virtue of which all free charges are integral multiple of basic unit of charge of e-. Thus, charge q = ne Cause of quantization is that only integral number of electrons can be transferred from one body to another. Page 7 Q. What is the charge on body which carries 20 excess e-? Sol: n = 20, q = -1.6 x 10^-19 C As q = ne, q = 20 x (-1.6 x 10^-19) = -3.2 x 10^-18 C Q. How many electrons constitute 1C of charge? Sol: n = ?, q = 1 C, e = -1.6 x 10^-19 C n = q/e = 1 / (1.6 x 10^-19) ≈ 10^19 electrons. Conservation of charge - Total charge of isolated system always remains constant or conserved. Thus, it is not possible to create or destroy net charge carried by any isolated system. However, charges can be created or destroyed in equal & unlike pairs only. Comparing charge & Mass Electric charge (q) / Mass q may be +ive, -ive, 0 / Mass is always +ive. q does not depend on velocity / Mass increase with velocity: m = m0 / sqrt(1 - v^2/c^2). Charge is quantised / quantisation of mass is yet to establish. q always conserved / mass not conserved, can be changed to energy. force b/w charges can be attractive or repulsive / gravitational force b/w masses is always attractive. Page 8 Coulomb’s Law -> quantitative statement about force b/w two point charges. Acc to it, Force b/w any two point charges is directly proportional to product of the charges & inversely proportional to square of distances b/w them. i.e. F ∝ (q1 q2) / r^2 q1, q2 are 2 point charges separated in vacuum by distance r. F = k |q1||q2| / r^2, k is electrostatic force constant. k depends on nature of medium separating charges, & on system of units. In air/vacuum, k = 9 x 10^9 Nm^2/C^2 We write k = 1 / (4πε0) = 9 x 10^9 Nm^2/C^2 ε0 is called absolute electrical permittivity of free space. F = (1 / 4πε0) * (|q1||q2| / r^2) ε0 = 1 / (k 4π) = 1 / (9 x 10^9 x 4 x π) ε0 = 8.85 x 10^-12 C^2 N^-1 m^-2 Page 9 Coulomb's law in vector form Let position vector of charge q1 be vector r1 = vector OA and q2 be vector r2 = vector OB. Vector leading from q1 to q2 is vector r12 = vector r2 - vector r1. Similarly, vector leading from q2 to q1 is vector r21 = vector r1 - vector r2. Direction of these: unit vector r12 = (vector r12 / |vector r12|) and unit vector r21 = (vector r21 / |vector r21|). If vector F12 = Force on q1 due to q2 and vector F21 = force on q2 due to q1. By coulomb’s law: vector F21 = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / AB^2] along AB. vector F21 = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / r12^2] * unit vector r12 = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / r12^2] * [vector r12 / |vector r12|]. vector F21 = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / r12^3] * (vector r2 - vector r1) = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / |vector r2 - vector r1|^3] * (vector r2 - vector r1). Similarly, vector F12 is force on q1 due to q2: vector F12 = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / r12^2] * unit vector r21 = - vector F21. Conclusion: Forces on q1 due to q2 is equal and opposite to force on q2 due to q1. Page 10 Unit of charge F = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / r^2]. For q1 = q2 = 1C and r = 1m, then F = 9 * 10^9 Newton. 9 * 10^9 = 9 * 10^9 * (1 * 1 / 1). Hence, one coulomb is that much charge which when placed in vacuum at a distance of 1 metre from an equal and similar charge, would repel it with force 9 * 10^9 Newton. Dielectric constant or relative electrical permittivity When charges are situated in medium other than free space, the force b/w them: F_m = [1 / (4 * pi * epsilon)] * [q1 * q2 / r^2], where epsilon is absolute electrical permittivity of intervening medium. In vacuum: F_0 = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / r^2]. On Dividing: F_0 / F_m = epsilon / epsilon_0 = epsilon_r or k. epsilon_r is relative electrical permittivity or dielectric constant. Hence Dielectric constant of a medium is ratio of force of interaction b/w two point charges in air/vacuum to force of attraction/repulsion b/w same 2 point charges held the same distance apart in medium. Page 11 Forces b/w Multiple charges: Principle of Superposition Acc to this, total force on any charge due to number of other charges at rest is the vector sum of all the forces on that charge due to other charges, taken one at a time. The force due to individual charges are unaffected by presence or absence of other charges. In general, the Force (vector F0) on a test charge q0 at position r0 due to all n discrete charges can be written as: vector F0 = vector F01 + vector F02 + vector F03 + ... + vector F0n. Acc to coulomb’s law: vector F01 = [1 / (4 * pi * epsilon_0)] * [q0 * q1 / r10^2] * unit vector r10. Putting these in equation: vector F0 = [1 / (4 * pi * epsilon_0)] * [ (q0 * q1 / r10^2) * unit vector r10 + (q0 * q2 / r20^2) * unit vector r20 + ... + (q0 * qn / rn0^2) * unit vector rn0 ]. vector F0 = [q0 / (4 * pi * epsilon_0)] * Sum(i=1 to n) [qi / ri0^2] * unit vector ri0. vector F0 = [q0 / (4 * pi * epsilon_0)] * Sum(i=1 to n) [qi / |vector r0 - vector ri|^3] * (vector r0 - vector ri). Page 12 Q. If body gives 10^9 electrons every second, how much time is required to get a total charge of 1C from it? Sol: n = 10^9, e = 1.6 * 10^-19 C, q = 1C. Charge given per sec = n * e = 10^9 * 1.6 * 10^-19 = 1.6 * 10^-10 C. Time required = (1 C / 1.6 * 10^-10) sec = 6.25 * 10^9 sec = approximately 198.18 years. Q. 3 charges q1, q2, q3 each equal to q placed at vertices of equilateral triangle. What is force on charge Q placed at centroid? Sol: In triangle ADB: AD = AB * cos(30) = (l * sqrt(3) / 2). Distance AO = (2/3) * AD = (2/3) * (l * sqrt(3) / 2) = l / sqrt(3). Force on Q at O due to q1=q at A: F1 = [1 / (4 * pi * epsilon_0)] * [Qq / AO^2] = [1 / (4 * pi * epsilon_0)] * [Qq / (l / sqrt(3))^2] = [3Qq / (4 * pi * epsilon_0 * l^2)] along AO. Similarly F2 along BO, F3 along CO. Angle b/w F2 and F3 = 120 degree. By parallelogram law of addition, resultant of F2 and F3 = [3Qq / (4 * pi * epsilon_0 * l^2)] along OA (opposite of F1). Total force on Q = [3Qq / (4 * pi * epsilon_0 * l^2)] - [3Qq / (4 * pi * epsilon_0 * l^2)] = 0. Page 13 Q. Consider 2 equal +ive charge each of 2 micro-Coulomb interact with 3rd +ive charge of 3 micro-Coulomb as shown. Calculate force of 3 micro-Coulomb charge. Sol: OA = OB = 3m, OP = 4m. AP = BP = sqrt(3^2 + 4^2) = 5m. Acc to coulomb’s law, Force on charge at P due to charge at A: F1 = [1 / (4 * pi * epsilon_0)] * [q1 * q2 / AP^2] = [9 * 10^9 * 2 * 10^-6 * 3 * 10^-6 / 5^2] = (54 / 25) * 10^-3 N = 2.16 * 10^-3 Newton. F1 is along PA'. Similarly F2 = F1 is along PB'. On resolving F1 and F2 we find sin component cancel out. Net F on 3 micro-Coulomb = 2 * F1 * cos(theta) = 2 * 2.16 * 10^-3 * (4 / 5) = 3.5 * 10^-3 Newton along Px. Page 14 Electric field — Its the space around charge in which electrostatic force of attraction or repulsion due to charge can be experienced. The charge Q which is producing field is called Source charge and the charge q0 which test the effect of source charge is called test charge. Electric field intensity — Its the strength of electric field at a given point. It is defined as force experienced by unit +ive charge placed at a Point. If vector F is the force experienced by unit +ive test charge q0 at any point vector r, then electric field intensity at this point is given by: vector E(r) = vector F(r) / q0. SI unit N/C. Its a vector. Direction of vector E is same as direction of vector F. i.e. vector E is along direction in which test charge q0 would tend to move if free to do so. The test charge q0 may have its own field and it may influence. So we take: vector E(r) = Limit (q0 -> 0) [vector F(r) / q0]. On account of discrete nature of charge, min value of q0 is 1.6 * 10^-19 C. q0 cannot be zero. Page 15 Electric Field intensity due to point charge vector F = [1 / (4 * pi * epsilon_0)] * [Q * q0 / r^2] * unit vector r0. Lets we have to calculate Electric field intensity at point P due to point charge Q at O. Imagine a small positive test charge q0 at P. Acc to Coulomb's law, Force on q0 at P: vector F = [1 / (4 * pi * epsilon_0)] * [Q * q0 / r0^2] * unit vector r0, where unit vector r0 is directed from Q to P. As vector E = vector F / q0, therefore vector E = [1 / (4 * pi * epsilon_0)] * [Q / r^2] * unit vector r0. Here Q is at Origin. Electric field vector around +q are directed outwards. Now: let +Q be at A, vector OA = vector ri. q0 be at P, vector OP = vector r0. Acc to coulomb's law, F on q0 at P: vector F = [1 / (4 * pi * epsilon_0)] * [Q * q0 / (ri0)^2] * unit vector ri0 = [1 / (4 * pi * epsilon_0)] * [Q * q0 / ri0^3] * vector ri0. vector F = [1 / (4 * pi * epsilon_0)] * [Q * q0 / |vector r0 - vector ri|^3] * (vector r0 - vector ri). Where (vector r0 - vector ri) = vector ri0 = vector AP. As Electric field = vector F / q0, therefore vector E = [1 / (4 * pi * epsilon_0)] * [Q * (vector r0 - vector ri) / |vector r0 - vector ri|^3]. Page 16 Electric field intensity due to group of charges It is the vector sum of vector E due to individual charges. For this we first calculate vector E due to given individual charge, as if other charges were not there and then add them vectorially. vector E = vector E1 + vector E2 + vector E3 + ... + vector En. vector E = [1 / (4 * pi * epsilon_0)] * Sum(i=1 to n) [Qi / ri^2] * unit vector ri. In vector form: vector E = [1 / (4 * pi * epsilon_0)] * Sum(i=1 to n) [Qi / |vector r0 - vector ri|^3] * (vector r0 - vector ri). Electric field intensity due to continuous charge distribution If lambda(ri) is linear charge density, than Force on +q0 at P(r0) due to this: vector F = [q0 / (4 * pi * epsilon_0)] * Integral [ (lambda(ri) * dl) / |vector r0 - vector ri|^3 ] * (vector r0 - vector ri). Therefore vector E = [1 / (4 * pi * epsilon_0)] * Integral [ (lambda(ri) * dl) / |vector r0 - vector ri|^3 ] * (vector r0 - vector ri). Similarly if sigma is surface charge density, vector E = [1 / (4 * pi * epsilon_0)] * Integral [ (sigma(ri) * ds) / |vector r0 - vector ri|^3 ] * (vector r0 - vector ri). Page 17 Physical significance of Electric field The E.F. around a charge gives us the force a unit +ive test charge would experience, when placed at that point. The field is independent of test charge that we place at a point to determine E.F. E.F. intensity can be calculated, vector F = q0 * vector E. Q. An electron falls through distance of 1.5 cm in uniform electric field 2 * 10^4 N/C. The direction of field is reversed and a proton falls through same distance. Compute time in each case. Sol: a) field is upward. so electron experience downward force. F = q * E. Acceleration = mass * a = q * E -> a = q * E / m. initial velocity u = 0, by equation of motion: s = u*t + (1/2) * a * t^2. we get t = sqrt(2 * s / a) = sqrt(2 * s * m / q * E). t_e (electron) = sqrt( (2 * 1.5 * 10^-2 * 9.11 * 10^-31) / (1.6 * 10^-19 * 2 * 10^4) ) = 2.9 * 10^-9 sec. b) mass of proton = 1.67 * 10^-27 kg. t_p (proton) = sqrt( (2 * 1.5 * 10^-2 * 1.67 * 10^-27) / (1.6 * 10^-19 * 2 * 10^4) ) = 1.3 * 10^-7 sec. Conclusion: Hence heavier particle (proton) take more time to fall same distance. Page 18 Electric field lines — path, straight or curved in E.F., such that tangent to it at any point gives the direction of E.F. intensity at that point. In fact, it is path along which unit positive charge actually moves in field, if free to do so. In figure, AB is electrostatic line of force. The tangent to this curve gives direction of E.F. Properties of E.F. lines They are continuous, starts from +ive and end at -ive charge. NO Electric lines of force exist inside charged body. Thus electrostatic field lines do not form continuous closed loops. Tangent to E.F. lines gives direction of E.F. intensity. No two E.F. lines can intersect each other. Always Normal to surface of conductor. exert lateral pressure on account of repulsion b/w like charges. E.F. intensity at a point is equal to number of field lines crossing normally a unit area around that point. Page 19 Electric field intensity -> number of electric field lines crossing normally a unit area around that point. E proportional to 1 / r^2 , i.e. Intensity is more near the charge and vice-versa. Electric Dipole — pair of equal and opposite point charges separated by some small distance. Total charge of dipole: -q + q = 0. Atom or molecules of ammonia, water, alcohol, CO2, HCl etc, are examples of electric dipole, because here centre of +ive and -ive charge distributions are separated by some small distance. Dipole moment (vector P) is a measure of strength of electric dipole. It is vector quantity whose magnitude is equal to product of magnitude of either charge and distance b/w them. |P| = q * (2a). By convention: Direction of vector P is from negative to positive. SI unit: C-m (Coulomb-meter). Page 20 Physical significance of electric dipole Study of electric dipole is important for electrical phenomena in matter. If centre of mass of +ive (proton) and centre of mass of -ive (electron) in an atom/molecule coincides, the molecule behaves as Non-polar molecule. If they do not coincide, then molecule is polar and it has some dipole moment. In absence of external electric field, dipole moments of different molecules are randomly arranged — So net dipole moment = 0. When external E.F. is applied, polar molecules tend to align themselves along field and some net dipole moment develops. The matter is said to have been polarised. Dipole field — Space around dipole in which electric effect of dipole can be experienced. Though the total charge of electric dipole is zero, the field of electric dipole is not zero because both charges are separated by some distance. However at larger distance i.e. r >> 2a, the field due to +q and -q largely cancel out. E proportional to 1 / r^3. Page 21 Field intensity on axial line of electric dipole We have to calculate vector E at point P on axial line of dipole at distance r from centre O of dipole. If vector E1 is E.F. at P due to charge -q at A and vector E2 due to +q at B. E1 = [1 / (4 * pi * epsilon_0)] * [q / (r + a)^2] along PA. E2 = [1 / (4 * pi * epsilon_0)] * [q / (r - a)^2] along BP. As vector E1 and vector E2 are collinear vectors acting in opposite directions and |E2| > |E1|, resultant E at P will be difference b/w two along BP. |E| = |E2| - |E1| = [q / (4 * pi * epsilon_0)] * [1 / (r - a)^2 - 1 / (r + a)^2]. |E| = [q / (4 * pi * epsilon_0)] * [ ( (r + a)^2 - (r - a)^2 ) / (r^2 - a^2)^2 ]. |E| = [q / (4 * pi * epsilon_0)] * [ 4 * a * r / (r^2 - a^2)^2 ]. But q * 2a = |P|. |E| = [1 / (4 * pi * epsilon_0)] * [ 2 * |P| * r / (r^2 - a^2)^2 ]. If dipole is short, 2a << r, then |E| = [1 / (4 * pi * epsilon_0)] * [ 2 * |P| / r^3 ]. E proportional to 1 / r^3 along BP produced. Page 22 Electric field intensity on Equatorial line of electric dipole We have to find vector E at P due to dipole. If E1 is electric intensity at P due to charge -q at A and E2 due to B +q at B. E1 = [1 / (4 * pi * epsilon_0)] * [q / (r^2 + a^2)] represented by vector PC. E2 = [1 / (4 * pi * epsilon_0)] * [q / (r^2 + a^2)] represented by vector PD. On resolving E1 and E2, we get E1cos(theta) along PR (parallel to BA) and E1sin(theta) along PE (perpendicular to BA). Similarly E2cos(theta) along PR and E2sin(theta) along PF (opposite to PE). E1sin(theta) and E2sin(theta) cancel out. Resultant |E| = E1cos(theta) + E2cos(theta) = 2 * E1 * cos(theta). |E| = [2 / (4 * pi * epsilon_0)] * [q / (r^2 + a^2)] * [a / sqrt(r^2 + a^2)]. |E| = [q * 2a / (4 * pi * epsilon_0 * (r^2 + a^2)^(3/2))]. But q * 2a = |P|. vector E = - vector P / [4 * pi * epsilon_0 * (r^2 + a^2)^(3/2)] along PR parallel to BA. If 2a << r, |E| = |P| / (4 * pi * epsilon_0 * r^3). E proportional to 1 / r^3. Page 23 Comparison: Due to Dipole: E proportional to 1 / r^3. Due to single charge: E proportional to 1 / r^2. E.F. due to short Electric Dipole at any point K We have to calculate E.F. intensity at K (dist r, angle theta). Dipole AB with dipole moment vector P can be resolved in 2 components: P*cos(theta) along A1B1. P*sin(theta) along A2B2 (perpendicular to A1B1). vector E1 at K on axial line of A1B1: |E1| = [2 * P * cos(theta)] / (4 * pi * epsilon_0 * r^3). vector E2 at K on equatorial line of A2B2: |E2| = [P * sin(theta)] / (4 * pi * epsilon_0 * r^3). Resultant |E| = sqrt(E1^2 + E2^2). |E| = [P / (4 * pi * epsilon_0 * r^3)] * sqrt(4 * cos^2(theta) + sin^2(theta)). Final Master equation: |E| = [P * sqrt(3 * cos^2(theta) + 1)] / (4 * pi * epsilon_0 * r^3). Direction: tan(alpha) = (1/2) * tan(theta). Page 24 Electric field intensity at any point on axis of a uniformly charged ring Let radius of loop be 'a', carrying total charge +q uniformly distributed. We have to determine vector E at P due to this loop. Let length of element AB = dl and C be the centre of element. Charge on AB is dq = (q / 2pia) * dl. E.F. at P due to charge element AB is dE. |dE| = [1 / (4 * pi * epsilon_0)] * [dq / (r^2 + a^2)]. Resolving dE: Notice that, as loop is considered to be made of large no. of pairs of diametrically opposite elements. Sum of dEsin(theta) = 0 as sin components cancel out. Hence resultant vector E at P is Sum(dE * cos(theta)). cos(theta) = r / sqrt(r^2 + a^2). |E| = Integral [ (1 / 4piepsilon_0) * (qdl / 2pia) * (r / (r^2 + a^2)^(3/2)) ]. |E| = [q * r / (4 * pi * epsilon_0 * (r^2 + a^2)^(3/2)) ] along PX, axis of loop. Page 25 Electric dipole in uniform 2-D Electric field Let dipole AB be held in uniform external electric field vector E at an angle theta. Force on +q at A = qE along direction of vector E. Force on -q at B = qE opposite to vector E. Since vector E is uniform, net force on dipole (qE - qE) = 0. However forces are equal, unlike and parallel, acting at different points, they form a couple, which rotates dipole anticlockwise direction. Torque = Force * arm of couple = F * AC. AC = 2a * sin(theta). Torque (tau) = (qE) * 2a * sin(theta) = (q * 2a) * E * sin(theta). tau = P * E * sin(theta). In vector form: vector tau = vector P x vector E. Special case: When vector P is along vector E (theta = 0), then tau = 0. This is Stable equilibrium. When theta = 180, tau = 0 (Unstable eq.). Max value of torque = P * E (at 90 degrees). Page 26 Potential energy of Dipole in a uniform electric field It is energy possessed by dipole by virtue of its particular position in electric field. Torque tau = P * E * sin(theta). Small work done in rotating dipole through small angle d(theta): dW = tau * d(theta) = P * E * sin(theta) * d(theta). Total work done in rotating from theta1 to theta2: W = Integral [P * E * sin(theta) * d(theta)] = P * E * [-cos(theta)] from theta1 to theta2. W = -P * E * [cos(theta2) - cos(theta1)]. Potential energy U = W = -P * E * (cos(theta2) - cos(theta1)). Particular cases: If dipole is initially aligned (theta1 = 0) and we set it to angle theta: W = P * E * (1 - cos(theta)). If initially at 90 degrees (theta1 = 90) and set to theta: W = -P * E * cos(theta). U = - vector P . vector E. U is scalar quantity, measured in Joule. Page 27 Q. Two charges +/- 10 micro-Coulomb are placed 5.0 mm apart. Determine E.F. at point P and Q. Sol: q = 10 micro-Coulomb = 10^-5 C. 2a = 5 mm = 5 * 10^-3 m. |P| = q * 2a = 10^-5 * 5 * 10^-3 = 5 * 10^-8 C-m. a) at point P, on axial line of dipole (r = 15 cm = 0.15 m): E1 = [1 / (4 * pi * epsilon_0)] * [2 * |P| / r^3] (as a << r). E1 = (9 * 10^9 * 2 * 5 * 10^-8) / (0.15)^3 = 2.67 * 10^5 N/C. b) at Q lies on equatorial line of dipole: E2 = [1 / (4 * pi * epsilon_0)] * [|P| / r^3] = (1/2) * E1. E2 = (1/2) * 2.67 * 10^5 = 1.33 * 10^5 N/C along a line parallel to BA. Page 28 Electric flux over an area is an electric field represents the total number of electric field lines crossing this area. If we place small planar element of area dS normal to vector E, the number of E.F. lines crossing this area element is proportional to E(dS). If we tilt area element by angle theta, the number of E.F. lines crossing area will be smaller. Electric flux (phi) = vector E . vector dS = E * dS * cos(theta). For closed surface: phi = Integral [E * dS * cos(theta)]. phi is a scalar quantity. When theta = 0, flux is max. When theta = 90, flux = 0. If theta > 90, flux is negative. Page 29 Gauss's Theorem in Electrostatics Total electric flux over closed surface S in vacuum is (1 / epsilon_0) times the total charge (Q) contained inside S. phi_E = Closed Integral [vector E . vector dS] = Q / epsilon_0. Location of charge Q inside S does not affect value of surface integral. Proof (for spherically symmetric surface): Let +q charge situated at centre O of sphere of radius r. Acc to coulomb's law, E.F. at any point P on surface of sphere is: vector E = [q / (4 * pi * epsilon_0 * r^2)] * unit vector r. Electric flux over area element dS: d(phi) = vector E . vector dS = E * dS (since angle is 0). Integrating over closed surface area: phi_E = Integral [ (q / 4piepsilon_0r^2) * dS ] = (q / 4piepsilon_0r^2) * (4 * pi * r^2). phi_E = q / epsilon_0. Page 30 If charges q1, q2 ... qn lies inside surface, then by principle of superposition: phi_E = (q1 + q2 + ... + qn) / epsilon_0 = Q / epsilon_0. Where Q = algebraic sum of all charges inside. If medium surrounding charge has a dielectric constant k, then phi_E = Q / (k * epsilon_0). If Q = 0, then phi = 0. Application: Coulomb's law from Gauss's theorem Direction of vector E due to +q at O and area vector are same (theta = 0). Acc to Gauss theorem: Closed Integral [E * dS] = q / epsilon_0. E * (4 * pi * r^2) = q / epsilon_0. E = q / (4 * pi * epsilon_0 * r^2). This is E.F. If +q0 charge be placed at P, then force on q0 would be: F = E * q0 = (q * q0) / (4 * pi * epsilon_0 * r^2) => Coulomb's law. Page 31 Applications of Gauss's Theorem Field due to infinitely long straight uniformly charged wire. To calculate field due to this infinite wire at P, we consider pair P1 and P2 equidistance from arbitrary origin O. Total field at any P is radial. Consider right circular cylinder of radius r and length l with infinitely long line of charge as its axis. vector E and unit vector n along outward normal to curved surface are in same direction (theta = 0). E. flux over curved surface of cylinder = Integral [E * dS] = E * (2 * pi * r * l). On the ends, angle b/w vector E and unit vector n is 90, so No flux. Total flux = E * (2 * pi * r * l). Charge enclosed in whole cylinder = linear charge density * length = lambda * l. Acc to gauss: phi = q / epsilon_0. E * (2 * pi * r * l) = (lambda * l) / epsilon_0. E = lambda / (2 * pi * epsilon_0 * r). E proportional to 1 / r. Page 32 Electric field due to Uniformly charged spherical shell R is radius of uniformly charged sphere centre O. Consider imaginary sphere S with radius r and centre O to calculate E.F. at P. Acc to Gauss theorem: Closed Integral [E * dS] = q / epsilon_0. E * (4 * pi * r^2) = q / epsilon_0. E = q / (4 * pi * epsilon_0 * r^2). This is E.F. produced by charge q placed at O. Hence, for point outside spherical shell, field due to uniformly charged shell is as if entire shell is concentrated at the centre of the shell. At a point on surface of shell (r = R): E = q / (4 * pi * epsilon_0 * R^2). If sigma is surface charge density, then q = 4 * pi * R^2 * sigma. E = sigma / epsilon_0 = constant. Field inside shell: Let radius of S2 be r < R. As charge inside shell is zero, i.e. Gauss surface enclose no charge. E * (4 * pi * r^2) = 0 / epsilon_0 = 0. E = 0 for r < R. Hence, field due to uniformly charged shell is zero at all points inside the shell. Page 33 E. field intensity due to Non-conducting Charged Solid sphere E.F. at P (OP = r). Imagine a sphere S, centre O, radius r which acts as Gaussian surface S. Closed Integral [E * dS] = q' / epsilon. where epsilon is electrical permittivity of material of insulating sphere. E * (4 * pi * r^2) = q' / epsilon. Now, charge inside S i.e. q' = Volume of S * volume density of charge. q' = (4/3) * pi * r^3 * rho. E = ( (4/3) * pi * r^3 * rho ) / ( 4 * pi * r^2 * epsilon ) = (rho * r) / (3 * epsilon). E proportional to r. i.e. E.F. inside non-conducting charged solid sphere is directly proportional to r. At centre, r = 0 so E = 0. At surface r = R, E = (rho * R) / (3 * epsilon) => maximum. Outside sphere, E proportional to 1 / r^2. Graphs: Due to Uniformly charged Spherical Shell: E=0 until r=R, then E decreases as 1/r^2. Due to Non-conducting charged solid sphere: E increases linearly until r=R, then decreases as 1/r^2. Page 34 E.F. intensity due to Thin infinite plane sheet of charge By symmetry, we find vector E on both side of sheet must be perpendicular to plane of sheet. let imagine a cylinder of cross-section dS around P and length 2r, piercing through sheet. At two cylindrical edges P and Q, vector E and outward normal unit vector n are parallel to each other. phi over edges = 2 * E * dS. On curved surface vector E and unit vector n are perpendicular, so phi = 0. Total charge enclosed by cylinder = sigma * dS. Acc to Gauss's: phi_E = 2 * E * dS = q / epsilon_0 = (sigma * dS) / epsilon_0. E = sigma / (2 * epsilon_0). here E is independent of r. If sheet carry +ive charge (sigma > 0), vector E will be normal and outwards from sheet. If sheet carry -ive charge (sigma < 0), vector E will be normal and inwards. Note: we have supposed sheet to be of infinite size so 'edge effect' can be ignored. Total E = E1 + E2 = sigma / epsilon_0 (if we consider both sides). Page 35 Electric field intensity due to two thin infinite parallel sheets of charge Let sigma1, sigma2 be uniform charge density of A and B. E1 and E2 are E.F. due to A and B. E1 = sigma1 / (2 * epsilon_0) and E2 = sigma2 / (2 * epsilon_0). In region I (left of A): E_I = -E1 - E2 = - (sigma1 + sigma2) / (2 * epsilon_0). In region II (between A and B): E_II = E1 - E2 = (sigma1 - sigma2) / (2 * epsilon_0). In region III (right of B): E_III = E1 + E2 = (sigma1 + sigma2) / (2 * epsilon_0). Special case: if sigma1 = sigma and sigma2 = -sigma. E1 = sigma / (2 * epsilon_0) and E2 = sigma / (2 * epsilon_0). E_I = 0. E_III = 0. E_II = (sigma / 2epsilon_0) - (-sigma / 2epsilon_0) = sigma / epsilon_0 = constant. Page 36 Educational Values and Methods: Innovative Teaching: Includes real demonstration of experiments, video clips, digital practical lab for Physics, Chemistry and Maths. 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