Question

A hollow metal sphere has 5 cm and 11 cm inner and outer radii, respectively. The surface charge density on the inside surface is −100nC/m2. The surface charge density on the exterior surface is +100nC/m2.

A) What is the strength of the electric field at point 4 cm from the center?

Express your answer to three significant figures and include the appropriate units.

B) What is the direction of the electric field at point 4 cm from the center?

C) What is the strength of the electric field at point 8 cm from the center?

Express your answer as an integer and include the appropriate units.

D) What is the direction of the electric field at point 8 cm from the center?

E) What is the strength of the electric field at point 12 cm from the center?

Express your answer to three significant figures and include the appropriate units.

F) What is the direction of the electric field at point 12 cm from the center?

Answer 1

(A) Solution:

Electric field strength at a point 4 cm away from center is Zero.

Reason: If we apply, Gauss's law, the charge enclosed within the Gaussian sphere is zero. Therefore, electric field becomes zero.

(B) Solution:

As the electric field is zero at a point 4 cm away from center, we can not have the direction for electric field.

(C) Solution:

Applying Gauss's law for a point, 8 cm away from the center.

where R = 8 cm

where Q is the chage enclosed within the gaussian loop which can be calculated as follows

where and r_i are surface chare density and radius of inner surface.

Susbtituting this equation in Gauss's law

substituting the given values

Simplifying this we will get

(D) Solution:

We have the electric field at a point 8 cm away from the center is -2825 V/m.

Here negative sign indicates that, the electric field in inward (towards center of the spherical shell)

(E) Solution:

Applying Gauss's law at a point 12 cm away from the center.

where R = 12 cm

In order to calculate the electric field first we need to find the chage enclosed withing the gaussian loop.

Enclosed charge due to inner surface is given by

Enclosed charge due to inner surface is given by

therefore, total chage enclosed is given by

Now, susbtituting this value in gauss's law

Simplifying this we will get

(F) Solution:

We have the electric field at a point 12 cm away from the center is

Here, positive sign indicates that the field is outward (away from the center of the spherical shell)