Question

10. Two conducting spheres with radii r and v2r with corresponding charges -go and 2g are briefly connected by a thin copper wire. What is the change in the electrostatic potential energy of the system? Assume I>>r. 11. Two conducting spheres, one with radius r and charge go the other uncharged with a radius 3r, are briefly connected by a thin wire. If I >> r, find the change in the electrostatic potential energy of the system. 9-0

Answer 1

Initial electro static potential energy of the system is (PE)_initial = -q_0 times 2 q_0/4 pi epsilon_) l [because l > > r] (PE)_initial = -q^2_0/2 pi epsilon_0 l let q_1 and q_2 be the charge in left and right sphere after briefly connected by the thin copper wire. From conservation of charge q_1 + q_2 = -q_0 + 2 q_0 \r\n q_1 + q_2 = q_0 -(i) since both spheres are connect by the wire so electric potentials of each sphere will be same that Is from coulombs law q_1/4 pi epsilon_0 r = q_2/4 pi epsilon_0 squareroot 2 r q_2 = squareroot 2 q_1 now from equation (i) q_1 + squareroot q_1 = q_0 q_1 = q_0/1 + squareroot 2 = 0.414 q_0 now q_2 = squareroot 2 times 0.414 q_0 = 0.585 q_0 now final electrostatic potential energy of the system is \r\n (PE)_final = q_1 q_2/4 pi epsilon_0 l (PE)_final = 0.414 q_) times 0.585 q_0/4 pi epsilon_0 l