A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b and outer radius c. The hollow sphere has no net charge.
Take V = 0 as r-> infinity. Use the electric field for this system:
to calculate the potential V at the following values of r.
A. r = c (at the outer surface of the hollow sphere)
B. r= b (at the inner surface of the hollow sphere)
C. r = a (at the surface of the solid sphere)
D. r = 0 (at the center of the solid sphere)

Explanation:

Consider the diagram for the charges on the given sphere(check attachment).

The electric field at this point are

E(r) = 0 for r≤a.  Equation 1

E(r) = kq/r² for a<r<b.   Equation 2

E(r) = 0 for b<r<c.      Equation 3

E(r) = kq/r² for r>c.    Equation 4.

We know that electric potential is related to the electric field using

V = Ed

A. The potential at outer surface of the hollow sphere (at r=c) can be calculated as,

The electric field at this point is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Then,

Vc = (kQ / c²) × c

Vc = kQ / c

Then, Q has charges +q, -q and +q

Then, Q = q - q + q = q

V = kq / c

B. The potential at inner surface of the hollow sphere (at r=b) can be calculated as,

V = kQ/r

V = kQ / b,   since r = b

Then, Q = q

V = kq / b

C. At r = a

Then, from equation 1.

E(r) = 0 for r≤a.  Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Then,

V = Ed = 0 V

So the electric potential at the surface of the solid sphere is 0

D. At r = 0

Then, electric potential can be calculated using

V = kq / r

Then, r = 0

V = kq / 0

V → ∞

i believe the answer is a

more than, less

less than, more