Question

Two spherical conductors (Homework #3) Two spherical conductors of different radii (Ri and R:) are connected by along fine conducting wire. Let's assume RiR, the smaller sphere carries the total charge Q (with surface charge density ) and the larger sphere carries the total charge a (with surface charge density a) Remember that the potential is always constant on and in the conductor. Using this fact, show that the ratio of the charge is given by R Q, R b) Show that the ratio ofthe charge density is given by c) Remember that the electrie field on the surface of the conductor is given by E- Using d) Recall the demo and SC Lab of the Van de Graaff generator, connected to a volunteer student. this fact, explain why the electric field becomes very large near sharp point. Explain why hivher hairs were stretched out.

Answer 1

In electro static equilibrium potential of both spherical conductors should be same V_A = V_B R Q_2/R_2 = RQ_1/R_1 Q_1/Q_2 = R_1/R_2 because surface change density sigma = Q/4 pi R^2 R sigma_2 4 pi R^2_2/R_2 = R sigma_1 pi R^2_1/R_1 sigma_1/sigma_2 = R_2/R_1 because E = sigma/epsilon_0 and in part(c) we understood that sigma proportionalto 1/R at sharp point, radius of curvature is very small if R is small then sigma will be large & hence E will be large In van de-graff generator, due to electrostatic repulsion, his/her hairs were stretched out.