Question

A uniformly charged ball of radius a and charge -Q is at the center of a hollow metal shell with inner radius b and outer radius c. The hollow sphere has a netcharge of +2Q
a. determine the the electric field strength in the four regions r_<a, a<r, b_<r_<c, and r>c
b. draw a graph of E versus r from r=0 to r=2c

Answer 1

Use Gauss’s law which relates the electric flux through a surface and the charge enclosed within the surface to solve for the electric field at different regions due to a system consisting of a uniformly charged ball surrounded by hollow metal shell.

The expression for the Gauss law is,

Here, is the charge enclosed within the surface, and is the flux through the surface.

The expression for the electric flux is,

Combine both equates and rearrange for E.

According to the concept of spherical symmetry of charge distribution, the direction of electric field will be radially inward or outward.

Choose Gaussian surface at where the charge is negative so the direction of electric field is inward.

From the Gauss’s law

Here, is volume charge density of the charged metal ball.

The relation between the volume charge density, volume, and charge is given by

If a ball has radius and net charge on a ball is then Volume of ball is ,

Now the above equation changes

Here, is charge distributed by the ball.

Substitute for in the equation

In vector notation

Therefore, the electric field at is.

Now, we will choose the Gaussian surface to the region, and on applying Gauss law to this region, we get,

Therefore, the electric field at is.

Choose the Gaussian surface is in the region of, and on applying Gauss law to this region, we get,

Here, the net field inside the metal shell is zero, because there is no net charge in the conductor or metal, and thus the flux through this Gaussian surface is zero.

Therefore, the electric field at is.

Let us choose the Gaussian surface in the region of , and on applying Gauss law to this region, we get,

The net charge on the shell is,

The electric field strength in the region of is,

In vector notation

Therefore, the electric field at is.