A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separated by vacuum and the inner radius of the outer shell is 4.30 cm.
Calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

Part(c):

If we apply Gauss' law of electrostatics, then

answer and explanation;