Coulomb's Law in Vector Form

COULOMB'S LAW IN VECTOR FORM Vector form of coulomb's law: r Q_{1} Q_{2} By substituting in hat r_{12} equation (3), we get, ...(4) tilde F_{21} Let us consider that two like charges, Q_{1} and Q_{2} are placed somewhere in space. According to coulomb's law, the electrostatic forces act along the direction of the line joining the two charges. Since we have taken the charges of the same nature, there must be a repulsive force acting between them. The force on charge Q: due to charge Q_{2} is vec F_{12} . The position vector of charge Q_{1} as seen from charge Q_{2} is vec r_{12} = vec r 12 where r is the separation between the charges. By applying coulomb's law, we get the following: Also, the position vector is given by, vec r_{12} =r hat r 12 Therefore, force acting on charge due to charge is given by. vec F 12 = (k*Q_{1}*Q_{2})/(r ^ 2) hat r 12 (1) F = (k*Q_{3}*Q_{2})/(r ^ 2) The unit position vector is given by, By substituting hat P_{12} in equation (1), we get, vec F 12 = (k*Q_{1}*Q_{2})/(r ^ 2) vec r 12 ... (2) Similarly, the force on charge Q_{2} due to charge Q_{1} vec F_{21} . The position vector of charge Q_{2}*i seen from charge Q_{1} iS, vec r_{21} =r hat r 21 Therefore, the force acting on charge due to charge is given by, The unit position vector is given by. ... (3) vec F 21 = (k*Q_{1}*Q_{2})/(r ^ 2) hat r 21 hat r_{12} = r_{12}/r hat r_{12} = vec r 12 r vec F 12 = (k*Q_{1}*Q_{22})/(r ^ 2) * vec r 12 r For unlike charges, the forces become nature and hence, their directions get changed. Therefore, the force on Q_{1} due to Q_{2} is expressed as, vec F 12 = (kappa*Q_{1}*Q_{2})/(r ^ 3) vec r 21 and the force on Q_{2} 1 due to Q_{1} is expressed as, vec F 21 = (k*Q_{1}*Q_{2})/(r ^ 3) vec r 12 vec F 21 = (k*Q_{1}*Q_{2})/(r ^ 3) vec r 21 Only the formulas for the like charges are enough to remember because to obtain the formulas for the unlike charges, we just need to put the value of charges with sign. Since vec r_{12} =- vec r 21 what we get from equations (1) and (2) is, vec F_{12} =- vec F 21 . Thus, coulomb's law agrees with newton's third law of motion. Force On Charge In A Medium The force experienced by two charged particles when positioned in vacuum (or in air) varies from the force acting on the same pair of charged particles when situated within a medium. When charge is introduced into a neutral medium (where opposite charges are balanced), the negative charges within the medium are drawn towards q_{1} resulting in a symmetric distribution of negative charges around qv These negative charges are termed as 'induced charges'. a + q_{1} vec F 12 = (k*Q_{1}*Q_{2})/(r ^ 2) * vec r 12 r q_{1} When a + qz charge is positioned near q_{1} the induced negative charges experience an attractive force from q_{2} Consequently, the distribution of the negative charges becomes asymmetric. Net force on charge q_{1} is: vec F ne*t_{1} = vec F 12 + vec F induced Net force on charge q_{1} will be lesser than F_{12} which is the force on q_{1} due to q_{2} only. Case 1: When tasked with computing the force exerted solely by q_{2} on in a vacuum, then q_{1} F_{12} = 1/(4pi*epsilon_{0}) * (q_{1}*q_{2})/(r ^ 2) F_{12} Case 2: When tasked with determining the total force acting on q_{1} due to q_{2} in a medium with permittivity epsilon r 1 then vec F ne*t_{1} = vec F 12 + vec F inducet = 1 4m epsilon 0 epsilon r q 1 q 2 r^ 2 = F 12 epsilon r Where 6 represents the relative permittivity or dielectric constant of the medium. e F_{12} 6 notin T F ind Superposition of coulomb's force: The resultant electrostatic force on a point will be the vector sum of electrostatic forces due to individual point charges. For finding the net force on any charge, we have to find the forces by each charge present in the vicinity of it. Let us consider charge Q_{1} and analyses all the forces acting on it. The forces acting on Q_{1} are shown in the Figure. The net force acting on Q_{1} is the vector sum of all the electrostatic force acting on it, which is given by, vec F net = vec F 12 + vec F 13 + vec F 14 + vec F 15 The force applied by one charge does not affect the force by other charges. They have their individual effects, but the net force acting on the charge changes. overline r_{21} - Q_{2} Q_{3} r in Q_{1} Q_{4} overline r_{14} r_{13} bullet Q 5 vec F_{13} F_{14} Q_{1} vec F_{15} vec F_{12} Ex Five balls numbered 1 to 5 are suspended using separate threads. Pairs (1, 2), (2, 4), and (4, 1) show electrostatic attraction, while pairs (2, 3) and (4, 5) show repulsion. What should ball 1 therefore be? (A) Positively charged (B) Negatively charged (C) Neutral (D) made of metal Sol. Consider five balls as shown in the figure. The dotted arrow represents the repulsive force between two balls and the solid arrow represents the attractive force between two balls. It is given that the pairs (1, 2), (2, 4), and (4, 1) show electrostatic attraction, while pairs (2, 3) and (4, 5) show repulsion. 10- 3 4 To show electrostatic repulsion, two charges must be like Charges. Suppose that there is al repulsive force between balls 2 and ball 3, assuming that they have charge A. Now, for electrostatic attraction to take place between two charges, it is not necessary that both the charges should have opposite charges. The electrostatic attraction can also happen between a charge particle and a neutral particle recall the example of the glass rod attracting dry papers). Now, since ball 4and ball 5 arer repelled to each other, they should have the same charge. Therefore, Ball 4 cannot be neutral. On top of this, ball 2 and ball 4 are attracted to each other and since ball 4 cannot be neutral, it should have the opposite charge of type a. Let ball 4 have a charge of type B. Therefore, ball 1 gets attracted by ball 2 having charge of type A and also gets attracted by ball 4 having charge of type B. Hence, it is possible only if ball 1 is neutral. Therefore, option (c) is the correct answer. Ex. Two identical charges in vacuum are separated by a distance of r. The electrostatic force between them is given by F. If 75% of the charge is taken from one of the charges and given to the other, then the new force becomes F'. Find the ratio F/y_{s} (A) 1 (B) * 16/9 (C 16/7 (D 7/16 Sol. Initially, let there be two identical charges of magnitude Q separated by a distance of r. Therefore, according to coulomb's law, the magnitude of the electrostatic force is, ...(1) Now, if 75% (or 3/4 ) of one charge is given to the other, then those two charges become, and Q ^ 2 = (Q + (3Q)/4) = (7Q)/4 However, the separation between them remains the same. Therefore, the electrostatic force in between is, Thus, option (C) is the correct answer. F' = (K(Q/4)((7Q)/4))/(R ^ 2) Q_{1} = Q/4 F' = (k*Q_{1}*Q_{2})/(r ^ 2) F' = 7/16 * F F = (k * Q ^ 2)/(r ^ 2) F' = 7/16 * (K * Q ^ 2)/(R ^ 3) F/(F') = 16/7 Ex. Two charge particles, each having charge q and mass m, are apart with a distance of d from each other. If two particles are in equilibrium under the gravitational and electrostatic force, then find the ratio q/m (A) 10-8 (B) 10-10 (C) 108 (D) None of these Sol. According to the problem, both the particles are in equilibrium under the gravitational and electrostatic force. So, it can be said that the electrostatic force is balanced by the gravitational force. Therefore, F elvctrastatie =F gravitational Thus, option (B) is the correct answer q/m \approx sqrt(10 ^ - 2) \approx 10 ^ - 1 q/m = sqrt(alpha/k) \approx sqrt((10 ^ - 1)/(10 ^ 9)) (k * q ^ 2)/(d ^ 2) = (G * m ^ 2)/(d ^ 2) (q/m) ^ 2 = G/k Ex. Two identical point charges +Q are fixed in a gravity-free space at points (L, 0) and (-1, 0). Another particle with mass m and charge -q is placed at the origin. Now, this particle is displaced by a distance of y along the y-axis and then released. Show that this particle will execute oscillatory motion. Ay +0 (-L, 0) -4 (2,0) (-L, 0) 0 F (L, 0) Sol. The only way to prove the particle executes oscillatory motion is to prove that the particle executes SHM because oscillatory motion is a consequence of SHM. Now, if we are able to prove that the force on the particle of charge -q is restoring and is proportional to the displacement, then it will be enough to conclude that the charged particle executes SHM. Suppose the particle of charge -q is displaced along the y-axis by a distance of y as shown in the figure. If the force on -q by +Q is given by F, then the net force on the charge-q by both the charges +Q is given by, Also, this force is pointing position (0) of the charge-q. Now, towards the equilibrium F net =2F cos theta Therefore, the net force on the charge -q is restoring in nature. F NET =2( KQQ R^ 2 )( Y R ) From the figure, it is seen that r = sqrt(L ^ 2 + y ^ 2) Since the charge is displaced slightly y < < L Hence, F net =2( kQq L^ 3 )y [Since y < < L] Therefore, the net force on the charge -q is proportional to the displacement of the charge from its equilibrium position and hence, the charge will execute SHM with time period, F net =2( kQq (L^ 2 +y^ 2 )^ 3 2 )y F net =2( kQu nu^ 3 (1+ y^ 2 12 )^ 1 2 )y F NET =2( KQQ R^ 3 )Y T = 2pi * sqrt((m * L ^ 3)/(2kQq)) T = 2pi * sqrt(m/(2((8qq)/(k ^ 2)))) F net =2F cos theta +0 Forces Between Multiple Charges The mutual electric force between two charges is given by Coulomb's law. How to calculate the force on a charge where there are not one but several charges around? Consider a system of n stationary charges q1, q2, and qu... q_{a} in vacuum. What is the force on qi due to q 2,q 3 ...q n ? Coulomb's law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin? Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.