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Chapter 3
Phys 2180
Electric Flux and Gauss’s Law
1- Electric Flux
Figure (1)
Consider an electric field that is uniform in both magnitude and direction, as shown
in Figure 1. The total number of lines penetrating the surface is proportional to the
product E A. This product of the magnitude of the electric field E and surface area A
perpendicular to the field is called the electric flux ΦE.
ΦE = E A
From the SI units of E and A, we see that ΦE 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 1
What is the electric flux through a sphere that has a radius of 1.0 m and carries a charge
of +1.0 µC at its center?
Solution
The magnitude of the electric field is giving by
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Chapter 3
Phys 2180
The field points radially outward and is therefore everywhere perpendicular to the surface
of the sphere. The flux through the sphere (whose surface area A = 4 πr2 = 12.6 m2) is
thus
If the surface under consideration is not perpendicular to the field, as show in Figure
2, the electric flux ΦE given by
ΦE = E A Cos θ
Figure (2)
From this equation, we see that the flux through a surface of fixed area A has a maximum
value EA when the surface is perpendicular to the field (when the normal to the surface is
parallel to the field, that is, θ = 0° in Figure 2); the flux is zero when the surface is
parallel to the field (when the normal to the surface is perpendicular to the field, that is, θ
= 90°).
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Chapter 3
Phys 2180
2- Gauss’s Law
In this section we describe a general relationship between the net electric flux
through a closed surface (often called a gaussian surface) and the charge enclosed by the
surface. This relationship, known as Gauss’s law, is of fundamental importance in the
study of electric fields.
Figure (3)
Let us again consider a positive point charge q located at the center of a sphere of radius
r, as shown in Figure 3. we know that the magnitude of the electric field everywhere on
the surface of the sphere is
E = Ke
q
r2
The field lines are directed radially outward and hence are perpendicular to the surface at
every point on the surface. That is, at each surface point, E is parallel to the vector ∆Ai
representing a local element of area ∆Ai surrounding the surface point. Therefore,
the net flux through the Gaussian surface is
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Chapter 3
Phys 2180
where we have moved E outside of the integral because E is constant over the surface
and given by E = K e
q
r2
Furthermore, because the surface is spherical,
Hence, the net flux through the Gaussian surface is
According to the superposition principle, which states that the electric field due to many
charges is the vector sum of the electric fields produced by the individual charges.
Therefore, we can express the flux through any closed surface as
Figure (4)
Figure 4 shows that the number of lines through S1 is equal to the number of lines
through the non-spherical surfaces S2 and S3. Therefore, we conclude that the net flux
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through any closed surface surrounding a point charge q is given by
q
εo
and is
independent of the shape of that surface.
Figure (5)
Consider the system of charges shown in Figure 5. The surface S surrounds only
one charge, q1; hence, the net flux through S is
q1
εo
. The flux through S due to charges
q2, q3, and q4 outside it is zero because each electric field line that enters S at one point
leaves it at another. The surface S\ surrounds charges q2 and q3; hence, the net flux
through it is
q 2 + q3
εo
. Finally, the net flux through surface S\\ is zero because there is no
charge inside this surface. That is, all the electric field lines that enter S\\ at one point
leave at another. Notice that charge q4 does not contribute to the net flux through any of
the surfaces because it is outside all of the surfaces.
Gauss’s law, which is a generalization of what we have just described, states that the
net flux through any closed surface is
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where qin represents the net charge inside the surface and E represents the electric field at
any point on the surface.
Example 2 (The Electric Field Due to a Point Charge)
Starting with Gauss’s law, calculate the electric field due to an isolated point charge q.
Solution:
We choose a spherical gaussian surface of radius r centered on the point charge, as shown
in Figure 6. The electric field due to a positive point charge is directed radially outward
by symmetry and is therefore normal to the surface at every point. Thus, E is parallel to
dA at each point. Therefore,
Gauss’s law gives
where we have used the fact that the surface area of a sphere is 4πr2. Now, we solve for
the electric field:
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Problems
(1) An electric field with a magnitude of 3.50 kN/C is applied along the x axis.
Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m
long assuming that (a) the plane is parallel to the yz plane; (b) the plane is parallel
to the xy plane; (c) the plane contains the y axis, and its normal makes an angle of
40.0° with the x axis.
Solution:
(2) Four closed surfaces, S1 through S4, together with the charges -2Q , +Q , and -Q
are sketched in the figure bellow. (The colored lines are the intersections of the
surfaces with the page.) Find the electric flux through each surface.
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Solution:
(3) A charge of 170 µC is at the center of a cube of edge 80.0 cm. (a) Find the total
flux through each face of the cube. (b) Find the flux through the whole surface of
the cube.
Solution:
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