A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b

Question

10 days ago

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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)

Physics

1 answer:

Softa [913] · Answer 1 · 2026-02-23T12:13:58+03:00

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

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 understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b, noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following 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

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.