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Why is the Qenclosed zero if there's a charge inside the shell?

A solid conducting sphere of radius a is placed inside a conducting shell which has an inner radius b and an outer radius c. There is a charge q1 on the sphere and a charge q2 on the shell.

Find the electric field at point P, where the distance from the center O to P is d, such that b<d<c.

There's a diagram which shows:

radius of solid conducting sphere = a

inner radius of conducting shell = b

outer radius of conducting shell = c

O is the center from which all radii are measured.

Flux = EA = Qencl/εnaught

A = 4∏r^2

E = Qenclosed/ (εnaught*4∏d^2)

I have the solution already, which is E=0, but I don't understand why the charge enclosed is 0. I understand that the charge on the shell spreads to the outer surface of that shell since it's a conductor, so that charge wouldn't be enclosed in the Gaussian sphere with radius d. I don't get why the charge on the solid sphere doesn't count, though.

In the problem before it, the question asked for the electric field at a point in between the solid sphere and the spherical shell, and the answer was kq1/d^2, which means that in the space between, the enclosed charge is q1.

## Homework Statement

A solid conducting sphere of radius a is placed inside a conducting shell which has an inner radius b and an outer radius c. There is a charge q1 on the sphere and a charge q2 on the shell.

Find the electric field at point P, where the distance from the center O to P is d, such that b<d<c.

There's a diagram which shows:

radius of solid conducting sphere = a

inner radius of conducting shell = b

outer radius of conducting shell = c

O is the center from which all radii are measured.

## Homework Equations

Flux = EA = Qencl/εnaught

A = 4∏r^2

## The Attempt at a Solution

E = Qenclosed/ (εnaught*4∏d^2)

I have the solution already, which is E=0, but I don't understand why the charge enclosed is 0. I understand that the charge on the shell spreads to the outer surface of that shell since it's a conductor, so that charge wouldn't be enclosed in the Gaussian sphere with radius d. I don't get why the charge on the solid sphere doesn't count, though.

In the problem before it, the question asked for the electric field at a point in between the solid sphere and the spherical shell, and the answer was kq1/d^2, which means that in the space between, the enclosed charge is q1.

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