# Electric field of concentric conducting spheres

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

## 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.

Last edited:

SammyS
Staff Emeritus
Homework Helper
Gold Member
The electric field in the conducting material of the shell itself is zero. That's a property of a conductor.

Use a spherical surface (for a sphere centered at O, having a radius between b and c) as your Gaussian surface. The E-field at all points on the Gaussian surface is zero. (The Gaussian surface lies entirely within conducting material.) Therefore, Gauss's Law tells us that the net charge enclosed within this sphere is zero.

There is a charge of q1 on the center sphere. The only other place (inside of our Gaussian surface) where charge can reside is on the inner surface of the spherical shell. Therefore, the charge on the inner surface of the shell (r = b) is -q1, making the charge enclosed equal to zero.

What is the net charge on the exterior surface of the shell?

gracy
Thanks. I understood that the field at points on the Gaussian surface is zero, but I don't understand where the -q1 comes from. The net charge on the exterior surface of the shell is +q2, and the net charge on the solid sphere is +q1. But why is it that at a point A between the inner shell surface and the outer solid sphere surface, the charge enclosed is +q1 (the charge on the solid sphere), but at a point P inside the shell (between outer and inner surfaces), all the sudden the charge enclosed is zero? Is it something like the +q1 from the solid sphere induces a charge of -q1 on the inner surface of the shell?

SammyS
Staff Emeritus
Homework Helper
Gold Member
The net charge on the shell (the sum of the charges on both inner & outer surfaces) is +q2. But, since the net charge enclosed in our Gaussian surface is zero, there is a charge of -q1 on the inner surface -- to cancel the +q1 charge on the central sphere.

The net charge on the shell is given by:
$q_{\text{Shell}}=q_\text{inner} + q_\text{outer}$

$+q_2= -q_1+ q_\text{outer}$​
Therefore,
$+q_2 +\, +q_1 =q_\text{outer}$​

gracy
Okay, thank you for helping. I just realized the gap in my logic which would be too confusing to explain...