Question

A hollow, conducting sphere with an outer radius of 0.260 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37 × 10−6 C/m2. A charge of -0.700 μC is now introduced into the cavity inside the sphere. a) What is the new charge density on the outside of the sphere? b) Calculate the strength of the electric field just outside the sphere. c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

Answer 1

a) Here, for the new charge density outside

New charge density = (6.37 * 10^-6 * 4pi * 0.260^2 - 0.70 *10^-6)/(4pi * 0.260^2)

New charge density = 5.55 *10^-6 C/m^2

new charge density is 5.55 *10^-6 C/m^2

b)

for the electric field

electric field = sigma/epsilon

electric field = 5.55 *10^-6/(8.854 *10^-12)

electric field = 6.27 *10^5 N/C

c)

just inside the inner surface

electric flux = Qin/epsilon

electric flux = - 0.70 *10^-6/(8.854 *10^-12)

electric flux = - 79060 N.m^2/C