<< Chapter < Page Chapter >> Page >

Strategy

Apply the Gauss’s law strategy given above, where we work out the enclosed charge integrals separately for cases inside and outside the sphere.

Solution

Since the given charge density function has only a radial dependence and no dependence on direction, we have a spherically symmetrical situation. Therefore, the magnitude of the electric field at any point is given above and the direction is radial. We just need to find the enclosed charge q enc , which depends on the location of the field point.

A note about symbols: We use r for locating charges in the charge distribution and r for locating the field point(s) at the Gaussian surface(s). The letter R is used for the radius of the charge distribution.

As charge density is not constant here, we need to integrate the charge density function over the volume enclosed by the Gaussian surface. Therefore, we set up the problem for charges in one spherical shell, say between r and r + d r , as shown in [link] . The volume of charges in the shell of infinitesimal width is equal to the product of the area of surface 4 π r 2 and the thickness d r . Multiplying the volume with the density at this location, which is a r n , gives the charge in the shell:

d q = a r n 4 π r 2 d r .
Figure shows four concentric circles. Starting from the smallest, their radii are labeled: r prime, r prime plus d r prime, R and r. The outermost circle is dotted and labeled Gaussian surface.
Spherical symmetry with non-uniform charge distribution. In this type of problem, we need four radii: R is the radius of the charge distribution, r is the radius of the Gaussian surface, r is the inner radius of the spherical shell, and r + d r is the outer radius of the spherical shell. The spherical shell is used to calculate the charge enclosed within the Gaussian surface. The range for r is from 0 to r for the field at a point inside the charge distribution and from 0 to R for the field at a point outside the charge distribution. If r > R , then the Gaussian surface encloses more volume than the charge distribution, but the additional volume does not contribute to q enc .

(a) Field at a point outside the charge distribution. In this case, the Gaussian surface, which contains the field point P , has a radius r that is greater than the radius R of the charge distribution, r > R . Therefore, all charges of the charge distribution are enclosed within the Gaussian surface. Note that the space between r = R and r = r is empty of charges and therefore does not contribute to the integral over the volume enclosed by the Gaussian surface:

q enc = d q = 0 R a r n 4 π r 2 d r = 4 π a n + 3 R n + 3 .

This is used in the general result for E out above to obtain the electric field at a point outside the charge distribution as

E out = [ a R n + 3 ε 0 ( n + 3 ) ] 1 r 2 r ^ ,

where r ^ is a unit vector in the direction from the origin to the field point at the Gaussian surface.

(b) Field at a point inside the charge distribution. The Gaussian surface is now buried inside the charge distribution, with r < R . Therefore, only those charges in the distribution that are within a distance r of the center of the spherical charge distribution count in r enc :

q enc = 0 r a r n 4 π r 2 d r = 4 π a n + 3 r n + 3 .

Now, using the general result above for E in , we find the electric field at a point that is a distance r from the center and lies within the charge distribution as

Practice Key Terms 3

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 2' conversation and receive update notifications?

Ask