Q. 54

Expert-verifiedFound in: Page 685

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

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

__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)

(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).

(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.

(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).

(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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