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Physics Lab
Gauss’s Law
I.
Electric Flux through Closed Surfaces


The electric flux through a set of imaginary surfaces, dAi , each with an electric field, Ei , can be
written as:






dE  E1  dA1  E2  dA2   and the total flux E  E  dA .


If the area is a closed surface, then the total flux is written as E  E  dA . This closed imaginary


surface is called a Gaussian surface. The area vectors at points on the Gaussian surface are chosen by
convention to point out of the enclosed region.
Remember also that the magnitude of the electric flux through a surface is proportional to the net
number of lines of force that pass through that surface.
The diagram to the right shows a Gaussian surface in the shape of a right circular
cylinder with radius R and height H. The top surface is labeled A, the cylindrical
side surface is labeled B, and the bottom surface is labeled C.
A.
surface
A
The Gaussian surface is in a uniform electric field of magnitude, Eo, which is
aligned with the axis of the cylinder.
1.
surface
B
Find the sign and magnitude of the flux through:
Surface A
Surface B
surface
C
Surface C
Eo
2.
Is the net (total) flux through the Gaussian surface positive, negative, or zero? Explain.
For the following three exercises, drawing the lines of force for the charge distributions will help you
answer the question.
B.
The Gaussian surface now encloses a negative charge, -Qo. (The uniform field Eo is removed.)
1.
Write the sign of the flux through
Surface A
Surface B
Surface C
- Qo
2.
C.
Is the net flux through the Gaussian surface positive, negative, or zero? Explain.
The Gaussian surface now encloses opposite charges of equal magnitude. (The charges are on the
axis of the cylinder and equidistant from the center.)
1.
Write the sign of the flux through
Surface A
Surface B
+Qo
Surface C
- Qo
2.
Is the net flux through the Gaussian surface positive, negative, or zero?
+Qo
D.
A positive charge is located above the Gaussian surface.
1.
Write the sign of the flux through
Surface A
2.
II.
Surface B
Surface C
Is the net flux through the Gaussian surface positive, negative, or zero? Explain.
Gauss’s Law
  q
Gauss’s Law ( E  E  dA  enclosed ) states that the electric flux through a Gaussian surface is directly
o
proportional to the net charge enclosed by the surface.

A.
Are your answers to section I, parts A, B, C, and D consistent with Gauss’s law? Explain.
2
B.
In section I, part D you determined the sign of the flux through the surfaces A, B, and C of the
cylindrical Gaussian surface. If A-10 N m2/C and C = +2 N m2/C, what is B?
C.
Find the values of the net flux through each of the cylindrical Gaussian surfaces below in terms of Qo
and constants.
1.
2.
3.
+5Qo
+Qo
4.
+Qo
- Qo
+100Qo
- Qo
- Qo
- 4Qo

D.

The three spherical Gaussian surfaces
to the right each enclose a charge
+Qo. In case C there is another
charge –6Qo outside the surface.
Consider the following conversation:


-6Qo
2r
r
+Qo
Case A
r
+Qo
Case B
+Qo
Case C
Student 1:
“Since each Gaussian
surface encloses the same
charge, the net flux through each must be the same.”
Student 2:
“Gauss’s law doesn’t apply here. The electric field at the Gaussian surface in case B is weaker
than in case A, because the surface is farther from the charge. Since the flux is proportional to
the electric field strength, the flux must also be smaller in case B.”
Student 3:
“I was comparing A and C. In C the charge outside changes the field over the whole surface.
The areas are the same, so the flux must be different.”
Do you agree with any of the students? Explain.
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III. Applications of Gauss’s law
A.
A large aluminum sphere of outside radius b has a spherical cavity
of radius a inside it. A charge of –2Qo is placed at the center of the
cavity.
b
a
1.
What is the magnitude and direction of the electric field in the
region between radii a and b? Explain.
2.
How much charge is located on the inside and outside surfaces of the aluminum sphere?
Explain your reasoning.
3.
a.
inside surface
b.
outside surface
-2Qo
If an additional amount of charge +5Qo is placed on the aluminum sphere, then how much
charge is on the inside and outside surface of the aluminum sphere?
a.
inside surface
b.
outside surface
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B.
Suppose the cavity in the previous problem was offset from the center of the
larger aluminum sphere.
1.
2.
3.
B.
What is the magnitude and direction of the electric field in the region
outside the cavity and inside the surface of the sphere?
-2Qo
How much charge is located on the cavity surface and on the outside
surface of the aluminum sphere?
a.
cavity surface
b.
outside surface
If an additional amount of charge +5Qo is placed on the aluminum sphere, then how much
charge is on the cavity surface and on the outside surface of the aluminum sphere?
a.
inside surface
b.
outside surface
A long cylindrical rod of radius R has a uniform charge density  distributed throughout its volume.
1.
What is the linear charge density ( = Q/L) of the rod? Express your result in terms of , R, and
constants. (Hint: Consider a length L of the rod and the amount of charge in that length.)
2.
Use Gauss’s Law to find the magnitude of the electric field at a distance r outside the rod, where
r is measured radially outward from the axis of the rod.
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3.
Use Gauss’s Law to find the magnitude of the electric field inside the rod at a radial distance r
from the axis of the rod.
4.
Extra Credit: Repeat parts 2 and 3 to this problem if the charge density of the rod is not
constant, but is a function of the distance from the axis of the rod,  = Cr, where C is a constant
and r is the radial distance from the axis of the rod.
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