A hollow metal sphere has inner radius a and outer radius b. When you measure th

Question

A hollow metal sphere has inner radius a and outer radius b. When you measure the electric field of the system at a point r > b you determine that the system has a total charge of +6q. You know that a point charge of +2q has been placed at the center of the cavity.

a. What is the electric field for r <= a.

b. What is the electric field for a < r < b.

c. What is the electric field for r >= b.

d. How much charge is on the interior surface of the sphere? How much on the exterior? What is the net charge of the sphere alone?

Explanation / Answer

here,

Use gauss's law. in each regionn, the flux is

fi = int(E*da) = 4*pi*R^2*E = Q/eo

Part A:
Since there is only the inner charge +Q inside the integral,
E = k*Q/R^2

Part B:
In electrostatics, there is no electric field inside the bulk of
a conductor.
E = 0.

Part C:
The total Charge inside the integration region, +3Q, so

E = 3*k*Q/R^2

Part D:
Since the electric field in part B was zero, the charge inside that integration region must be zero. This includes the charge in the middle and the charge on the inner surface. Therefore, the charge on the inner surface is –Q.

We know the total charge on the metal sphere is +2Q

this charge is divided between the inner and outer surfaces. Therefore, the charge on the outer surface is +3Q.

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