Names: _______KEY___________________________________________________ Group:___
Gauss’s Law Tutorial – Group Answers
Part A:
Draw a notional boundary around three objects:
Draw anything
Is there any particular shape for your boundary? Any shape will work as long as three
balls are enclosed
Part B:
How many fields are leaving the notional boundary? ___4____
Are there any field lines entering the notional boundary? ___no____
If the points are negative, how would that change your answer? Now there are 0 leaving
and 4 entering so the net number is -4.
How do the signs net # of fields lines passing through a boundary to the sign of the
charge enclosed by the boundary compare? Sign of charge=sign of net # of field lines
What happens to net # if field lines if the charge enclosed is doubled? Tripled? Explain.
Proportional (doubling charge doubles # of field lines)
How many field lines leave the new notional boundary? ____5______
How many field lines are entering? ___1_____ What is the net # of field lines? ___4___
Can you draw a boundary that encloses the charged ball and does not have the same #
of net field lines passing through it? No
What is the net # of filed lines passing through the picture in figure 5? ____0___
What is the net charge enclosed in this surface? 0
Describe the relationship between the net charge enclosed and the net # of field lines.
Directly proportional
Part C:
If you draw a boundary enclosing a 16µC charge, what is the net # of field lines passing
through the boundary? 16 What if the ball has a -16µC charge? -16
If you draw a boundary around two charges of equal mag but opposite sign, what is the
net # of field lines passing through that boundary? 0
Can you confirm your answers with the drawing? Yes- Just circle everything (4 in, 4 out
= 0 net)
Part D:
Why are area vectors not parallel to the surface? Can’t define just one – infinite
orientation. Only 2 perpendicular options…
What does the magnitude of the area vector tell you? Size of area
Draw an area vector for the surface that makes the electric flux through it +: rightward
Draw an area vector for the surface that makes the electric flux through it -: leftward
If we double the area of the surface, what happens to the number of field lines passing
through it? It doubles
If we double the strength of the electric field, what happens to the number of field lines
passing through the surface? It doubles
Is the electric flux through the surface in Fig. 7 the same as it was in Fig. 6? Why? No,
not as many lines pass through now (angle is different)
Two vectors have been drawn on Fig. 7 Label those vectors E and A. Explain why you
labeled the way you did. E is parallel to the field and A is perpendicular to the surface
What is the mag of the electric flux if the angle θ is 90®? 0 What if θ is 0®? max
Would you say that the electric flux is directly proportional to sinθ or cosθ? Why?
Follows a cosine curve
Part E:
What are the area vectors for the three figures in Fig. 8?
How does the E flux passing through the left and right face of the cube compare? equal
What is the electric flux passing through each of the other faces? 0 (A is perpendicular
to E)
What is the net electric flux passing through the surface? 0
How many area vectors are needed to describe surface (a) in Fig. 8? Why? Infinite (with
infinitesimal magnitude)
Draw field lines to represent the direction of the E field at the indicated points. Radially
out from center – so perpendicular to the surface
What is the angle between the electric field vector and the area vector at each point? 0
Does the mag. of the electric field change as you look at different points on the sphere?
No
Justify the individual mathematical steps in question 4.
Step 1: cosθ=1
Step 2: E is constant at surface
Solve for the flux in question 5. =1/4πε0 (Q/r2) * (4πr2)
=Q/ ε0
Look back to question 8 in part B. What is the proportionality constant that equates
electric flux through a surface to the charge enclosed by that surface? ε0
Draw field lines to represent the electric field around the point charge in Figure 11.
Is the angle, θ, constant across the surface of the cylinder? No
Does the mag of the electric field change as you look at different points on the cylinder?
Yes
Can the same mathematical steps be taken with the integral for this setup? Discuss the
difficulty in using the cylinder as a Gaussian surface here with a point charge. Steps
used with a sphere can’t be done here!
Part F:
Describe the electric field around a line of charge.
How is the field symmetric? Cylindrically symmetric around the wire
Choose a notional surface to enclose one meter of the line of charge
What is the flux through the two ends of your notional surface? 0 through ends. E
perpendicular to dA
What is the area of the rest of your notional surface? Area=2πrL
Use Gauss’ Law to relate the flux to the charge enclosed by the charge and solve for the
electric field.
Problems:
A pyramid has a square base of side a, and four faces which are equilateral triangles. A charge Q
is placed on the center of the base of the pyramid. What is the net flux of electric field
emerging from one of the triangular faces of the pyramid?
Construct an eight faced closed surface consisting of two pyramids with the charge at the
center. The total flux by Gauss’s law is just Qenc / ε0 . Since each face is identical, the flux
through each face is one eight the total flux or Q / 8ε 0 .
A very long conducting cylinder (length L and radius a) carrying a total charge +q is surrounded
by a thin conducting cylindrical shell (length L and radius b) with total charge –q, as shown in
the figure.
(a) Using Gauss’s Law, find an expression for the
direction and magnitude of the electric
field E for the region r < a.
(b) Similarly, find an expression for the direction
and magnitude of the electric field E for the region
a< r< b .
a) The electric field inside a conductor is always 0. There is no enclosed charge here (all
the charges are equally distributed on the outer surface of the inner cylinder) which is
another way of seeing how the electric field is 0.
b) Imagine a Gaussian cylinder of radius r and length L between the inner rod and the
outer shell. The Electric field will have radial symmetry and will be constant on this
surface (and parallel to the area vector for the curved surface of the cylinder. Flux will
be 0 through the caps of this Gaussian surface because the electric field and area
vectors are perpendicular.
𝑄
∮ 𝐸⃗ ∙ 𝑑𝑎 = ∮ 𝐸𝑑𝑎 = 𝐸 ∮ 𝑑𝑎 = 𝐸𝐴 = 𝐸2𝜋𝑟𝐿 By Gauss’s Law, this is equal to 𝑒𝑛𝑐⁄𝜀0
where the charge enclosed is the charge density times the length of the cylinder. This
means 𝐸2𝜋𝑟𝐿 =
𝜆𝐿
𝜀0
𝜆
So in this region, 𝐸 = 2𝜋𝜀
0𝑟
GRADING RUBRIC: 10 pts for completeness for the problems here (4 pages of worksheet), up to
10 points for the ponderable solution (5 pts per part of the question).