Gauss Law

2.2

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PUZZLER
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chapter

Gauss’s Law
Chapter Outline
24.1 Electric Flux
24.2 Gauss’s Law
24.3 Application of Gauss’s Law to
Charged Insulators

24.4 Conductors in Electrostatic

24.5 (Optional) Experimental
Verification of Gauss’s Law and
Coulomb’s Law

24.6 (Optional) Formal Derivation of
Gauss’s Law

Equilibrium

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CHAPTER 24

Gauss’s Law

I

Area = A

E

n the preceding chapter we showed how to use Coulomb’s law to calculate the
electric field generated by a given charge distribution. In this chapter, we describe Gauss’s law and an alternative procedure for calculating electric fields.
The law is based on the fact that the fundamental electrostatic force between point
charges exhibits aninverse-square behavior. Although a consequence of
Coulomb’s law, Gauss’s law is more convenient for calculating the electric fields of
highly symmetric charge distributions and makes possible useful qualitative reasoning when we are dealing with complicated problems.

24.1
Figure 24.1 Field lines representing a uniform electric field
11.6
penetrating a plane of area A perpendicular to thefield. The electric
flux E through this area is equal
to EA.

ELECTRIC FLUX

The concept of electric field lines is described qualitatively in Chapter 23. We now
use the concept of electric flux to treat electric field lines in a more quantitative
way.
Consider an electric field that is uniform in both magnitude and direction, as
shown in Figure 24.1. The field lines penetrate a rectangularsurface of area A,
which is perpendicular to the field. Recall from Section 23.6 that the number of
lines per unit area (in other words, the line density) is proportional to the magnitude of the electric field. Therefore, the total number of lines penetrating the surface is proportional to the product EA. This product of the magnitude of the electric field E and surface area A perpendicular to the fieldis called the electric flux
E (uppercase Greek phi):
E

(24.1)

EA

From the SI units of E and A, we see that E has units of newton – meters squared
per coulomb (N m2/C). Electric flux is proportional to the number of electric field lines penetrating some surface.

EXAMPLE 24.1

Flux Through a Sphere

What is the electric flux through a sphere that has a radius of
1.00 m and carries acharge of 1.00 C at its center?

perpendicular to the surface of the sphere. The flux through
the sphere (whose surface area A 4 r 2 12.6 m2 ) is thus
E

Solution The magnitude of the electric field 1.00 m from
this charge is given by Equation 23.4,
E

ke

q
r2

8.99

(8.99

10 9 N m2/C 2 )

1.00 10 6 C
(1.00 m )2

10 3 N/C

The field points radially outward and is thereforeeverywhere

(8.99

EA
1.13

10 3 N/C)(12.6 m2 )

10 5 N m2/C

Exercise

What would be the (a) electric field and (b) flux
through the sphere if it had a radius of 0.500 m?

Answer

(a) 3.60

10 4 N/C; (b) 1.13

10 5 N m2/C.

If the surface under consideration is not perpendicular to the field, the flux
through it must be less than that given by Equation 24.1. We can understandthis
by considering Figure 24.2, in which the normal to the surface of area A is at an
angle to the uniform electric field. Note that the number of lines that cross this
area A is equal to the number that cross the area A , which is a projection of area A
aligned perpendicular to the field. From Figure 24.2 we see that the two areas are
related by A
A cos . Because the flux through A...