PHYS208 Fundamentals of Physics II

Quiz 4 -- Gauss's Law

A solid nonconductive spherical ball of diameter 1.0 cm is uniformly charged such that is has an electric field of 20 kV/m at its surface.

[ball of charge]

1. How much charge is on the ball?
2. What is the magnitude of the electric field halfway to the center of the ball?
3. If the ball were made of a conductive material, how would the answers to 1) and 2) change?


Step 1: Understand the Geomtery:

The charge distribution is spherically symmetric, extending out to radius R.

Step 2: Understand the symmetry: The only variable on which E may depend is r, the distance from the center of the sphere of charge. There is no angular dependence because the charged object is spherically symmetric, i.e. it does not matter at which angle about the sphere you view it.

Similar arguments determine that the direction of E is radially away from the center of the sphere, perpendicular to the surface of the sphere at all locations.

Step 3: Construct the gaussian surface: Given the symmetry of the charge distribution and resulting electric field, the appropriate choice for gaussian surface is a sphere of radius r centered about the charged sphere.

For question 1 of this quiz, a gaussian surface should be chosen that has a radius that just encloses the entire ball.

gaussian surface just outside sphere
For question 2, the radius of the gaussian surface should be exactly one-half of the ball radius.

gaussian surface inside sphere

Step 4: Examine the gaussian surface:

The gaussian sphere consists of one simple surface (unlike the case of the gaussing cylinder or can consisting of a cylindrical tube and two flat end caps). Everywhere on the surface of the gaussian sphere the electric field is pointing in the same direction as the infinitesimal area vector. Also, the magnitude of E is constant everywhere on the gaussian surface.

Step 5: Evaluate the electric flux:

The electric flux for the gaussian surfaces above, both inside the charged sphere and outside is:

Evaluation of flux through spherical gaussian surface

Step 6: Evaluate the enclosed charge:

The evaluation of the charge enclosed by a sphere of radius r need not involve an integral for either question, since the charge density has been assumed to be constant; i.e. charge distributed uniformly. involves the integral of the density over the enclosed volume.

For question 1, the enclosed charge is the total charge of the ball, Q.

For question 2, the enclosed charge is just one-eighth that of the ball since the radius has been halved and the volume enclosed by the gaussian sphere is proportional to the cube of the radius. In general, for a gaussian sphere of radius r inside a ball of uniform charge Q having radius R, the enclosed charge will be given by:

charge enclosed by gaussian surface inside the ball

Step 7: Apply Gauss's law:

For question 1,
charge completely enclosed by gaussian sphere

So given ER = 20,000 V/m and R = 0.005 m, the final equation above may be used to determine that the total charge on the ball is 56 pC.

For question 2,

halfway to center of ball

Thus the electric field inside the ball depends linearly on the distance from the center. At a position halfway to the center, r/R = 1/2 and the electric field is thus one-half of the value at the ball's surface; that is, 10 kV/m.

For question 3, remember that the electric field must be zero at equilibrium inside a solid conductor. Also, for spherically symmetric charge distributions, in the region outside the charged object, the electric field is the same as if the charge were concentrated at a point at the center of the distribution. Thus for a charged conductor with the same electric field as a charged insulator at its surface, the total charge would be required to be the same. However the electric field would be zero, whereas for the uniformly-charged insulator the electric field depends linearly with distance from the center.

Last updated March 16, 1998.
Copyright George Watson, Univ. of Delaware, 1998.