Electric Charges and Fields

Electric Charges and Fields Problems Question1. A comb drawn through person’s hair on dry day causes 1022 electrons to the person’s hair and stick to the comb. Calculate the charge carried by the comb. Solution. Here n = 1022 e = 1.6 × 10-19 c q = ne q = 1022 × 1.6 × 10-19 q = 1.6 × 103 C As the comb has excess of electron, Therefore comb on charge = -1.6 × 10-3 c Question2. The electrostatic force of repulsion between two positively charged ions carrying equal charges is 3.7 × 10-9 N, when they are separated by a distance of 5 A0. How many electrons are missing from each ion? Solution. Here F = 3.7 × 10-9 N, r = 5 A0 = 5 × 10-10 m, q1 = q2 = q (say) As F = 3.7 × 10-9 = 9 × 109 Or q2 = = 10.28 × Or q = 3.2 × C Number of electrons missing from each ion is q = ne n = = = 2 Question3. A free pith- ball A of 8 g carries a positive charge of 5 × C. What must be the nature and magnitude of charge that should be given to a second pith ball B fixed 5 cm below the former ball so that that upper ball is stationary? Solution. The pith ball B must be positive charge i.e., of same nature as that of A, so that the upward force of repulsion balance the weight of pith ball A. When the pith ball A remains stationary, F = m1g 0r = m1g But m1 = 8 g = 8 × 10-3 kg q1 = 5 × 10-8 C r = 5 cm = 0.05 m 9 × 10-9 = 8 × 10-3 × 9.8 Or q2 = q2 = 4.36 × 10-7 C (Positive) Question4. A particle of mass m and carrying charges –q1 is moving around a charge +q2 along a circular path of radius r. Prove that the period of revolution of the charge –q1 about +q2 is given by T = Solution. Suppose charge –q1 moves around the charges +q2 with the speed v along the circular path of radius r. than Force of attraction between the two charges = Centripetal force = Or v = This period of revolution of charge –q1 around +q2 will be T = Or T = 2 T = Question5. Two particles, each are having a mass of 5 g and charge 1.0 × 10-7 C, stay in limiting equilibrium on a horizontal table with a separation of 10 cm between them. The coefficient of friction between each particle and the table is the same. Fine μ. Solution. Here q1 = q2 = 1.0 × 10-7 C, r = 10 cm = 0.10 m, m = 5 g = 5 × 10-3 kg, The mutual electrostatic force between the two particles is F = F = 9 × 109 F = 0.009 N The limiting force of friction between a particle and the a table is F = μ × mg F = μ × 5 × 10-7 × 9.8 F = 0.049 μ N As the two forces balance each other, therefore 0.049 μ = 0.009 μ= =0.18 question6 (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10-7 C? The radii of A and B are negligible compared to the distance of separation. Also compare this force with their mutual gravitational attraction if each weight 0.5 kg. (b) What is the force of repulsion if, (1) Each sphere is charged doubled the above amount, and the distance between them is halved. (2) The two sphere are placed in water? (Dielectric constant of water= 80). Solution. (a) Here q1 = q2 = 6.5 × 10-7 C, R = 50 cm = 0.50 m, Using Coulomb’s law, Fair = k Fair = 9 × 109 Fair = 1.5 × 10-2 N The mutual gravitational attraction, FG = G FG = 6.67 × 10-11 FG = 6.67 × 10-11 N Clearly, FG << Fair. (b) (1) When charge on each other sphere is doubled, and the distance between them is halved, the force of repulsion becomes, Fair’ = k Fair’ =16 k Fair’ = 16 × 1.5 × 10-2 Fair’ = 0.24 N (2) The force between two charges placed on a medium of dielectric constant k is given by. F = For water, k = 80 FWater = FWater = Fwater = 1.875 × 10-4 N Fwater = 1.9 × 10-4 N Question7. The similarly equally charged identical metal sphere A and B repel each other with a force of 2.0 × 10-5 N. A third identical uncharged sphere C is touched to A, then placed at the midpoint between A and B. Calculate the net electrostatic force on C. Solution. Let the charge on each of the spheres A and B is q. If the separation between A and B is r then electrostatic force between spheres A and B will be F = K = 2.0 × 10-5 N When sphere C is touched to A, the sphere charge q/2 each, because both are identical, Force on C due to A = K = K , along AC Force on C due to B = K = K , along BC Since, these forces act ion opposite direction, Therefore net force on C is F’ = K - K F’ = K =2.0 × 10-5 N, along BC. Question8. Two identical charges Q each are kept at a distance r from each other. A third charge q is placed on the line joining the above two charges such that all the three charges are in equilibrium. What is the magnitude, sign and position of the charge q? Solution. Suppose the three charges be placed in the manner The charges q will be in equilibrium if the force exerted on it by the charges at A and C are equal and opposite. = k Or x2 = (r-x)2 Or x = (r-x) Or x = Since the charge at A is repelled by the similar charge C so it will be in equilibrium if it is allracted by the charge q at B I.e., the sign of charge q should be opposite to that of charge Q. Force of repulsion between charges A and C = Force of attraction between charges Or = k Or q = Question9. Two point charges +4q and +e are fixed a distance a apart. Where should a third point charge q be placed on the line joining the two charges so that it may be in equilibrium? In which case the equilibrium will be stable and in which unstable? Solution. Suppose the third charges are placed. Let the charge q be positive. For equilibrium of charge +q, we must have force of repulsion F1 between +4e and +q = force of repulsion F2 between +e and +q. = k Or 4 (a- x)2 = x2 Or 2 (a- x) = x x = or 2a As the charge q is placed between +4e and +e, so only x = 2a/3 is possible. Hence for equilibrium, the charge q must be placed at a distance 2a/3 from the charge +4e. We have considered the charge q to be positive. If we displace it slightly toward charge e, from the equilibrium position then F1 will be decrease and F2 will increase and a net force (F2 – F1 ) will acts on q towards left i.e., toward the equilibrium position. Hence the equilibrium of positive q is stable. Now if we take charge q to be negative, the force F1 and F2 will be attractive The charges –q will still be in equilibrium at x = 2a/3. However if we displace charge –q slightly toward right then F1 will decrease and F2 will increase. A net force ( F2 –F1) will act on –q toward right i.e., away from the equilibrium position. So the equilibrium of the negative q will be the unstable. Question10. Two “free” point charges +4e and +e placed a distance ‘a ‘apart. Where should a third point charge q be placed between them such that the entire system may be in equilibrium? What should be the magnitude and sign of q? What type of a equilibrium will it be? Solution. Solution the charge are placed as As the charge +e exerts repulsion F on charge +4e so for the equilibrium of charge +4e, the charge –q must exert attraction F’ on +4e. This requires the charge q to be negative. For equilibrium of charge +e. F = F’ = Or q = For equilibrium of charge,-q Attraction F1 between +4e and –e = Attraction F2 between +e and –q = Or x2 = 4 (a-x)2 X = 2a / 3 Hence q = q = . q = The equilibrium of the negative charge q will be unstable.