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Q. 54

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Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
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Short Answer

A spherical shell has inner radius and outer radius . The shell contains total charge , uniformly distributed. The interior of the shell is empty of charge and matter.

a. Find the electric field strength outside the shell, .

b. Find the electric field strength in the interior of the shell, .

c. Find the electric field strength within the shell, .

d. Show that your solutions match at both the inner and outer boundaries

a.

b.

c.

d. We've shown that our solutions match at both the inner and outer boundaries

See the step by step solution

Step by Step Solution

Step 1 : Given information and Theory used 

Given :

Inner radius and outer radius of the spherical shell :

The shell contains total charge :

The interior of the shell is empty of charge and matter.

Theory used :

The quantity of electric field that passes through a closed surface is referred to as the electric flux. The electric flux through a surface is proportional to the charge inside the surface, according to Gauss's law, which is given by :

(1)

Step 2 : Finding the electric field strength outside the shell 

(a) The Gaussian surface encapsulated net charge for a point outside the sphere. The area of the enclosed surface is for every point outside the sphere at a distance greater than the radius of the sphere Rin.

In the gaussian surface, the enclosed charge .

To get the electric field outside the sphere, plug the equations of into equation (1).

Step 3 : Finding the electric field strength interior of the shell

(b) The electric field is zero at and within the sphere because the net charge inside a conductor is zero. That is, and the electric field depends on the amount of charge within the bulk, hence the electric field is zero.

Step 4 : Finding the electric field strength within the shell 

(c) Between , the volume charge density of the gaussian surface is :

Since the charge density is connected to the sphere it's computed by

The enclosed charge will be

Plug this expression of into equation (1).

Step 5 : Showing that our solutions match at both the inner and outer boundaries

(d) The electric field in section (c) will be

if the radius of the gaussian surface is the same for the inner

Because this is the same electric field as part (b), our responses at the boundary at are identical.

The electric field in component (c) will be

if the radius of the gaussian surface is the same for the outer radius

Because this is the same electric field as part (a), our responses at the boundary at are identical.

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