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Exam I Review.
RI-1. Three equal mass charges are released from rest at the positions shown on the xaxis. Which mass has the largest initial acceleration?
+Q
+Q
+
+
-Q
-
A
B
C
A: A
B: B
C: C
D: 2 of the masses have the same acceleration.
x
E: Some other answer.
Answer: B. The particle which feels the largest E-field due to it neighbors will have the
largest force. Since all the masses are equal, the largest force produces the largest
acceleration.
RI-2.
Four charges are arranged as shown, all a distance r from the origin. What is the
magnitude of the E-field at the origin, in units if kQ/r2 ?
y
+Q
+2Q
x
+2Q
-4Q
A: 1
B: 2
C: 3
D: 4
E: None of these.
Answer: None of these. The answer is 5. The magnitude of the E-field is 5kQ/r2.
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RI-3. A line of charge has length L and total charge Q. What is the relation between dx
and dQ?
A: dx = (Q/L) dQ
B: dQ = (L/Q) dx
C: dQ = (Q/L) dx
D: None
dE
point A

h
dx, dQ
0
x
L
At point A, what is the magnitude of the field dE due to the element of charge dQ?
k dQ
k dQ
A: 2
h  x2
B:
C:
h2  x2
k dQ
h2  x2
What is the correct expression for sin?
A:
x
h2  x2
h
B:
C:
h2  x2
h
h2  x2
x
D:
What is the correct expression for dEx?
k  Q / L  dx
A:
h
2
 x2

3/ 2
Answers: dQ = (Q/L) dx
dE x  dE sin  
k x  Q / L  dx
k h (Q / L)dx
B:
h2  x2
k dQ
h2  x2
k x  Q / L  dx
dE 
h
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2
 x2

h2  x2
C:
sin  
h
2
 x2

3/ 2
x
h2  x2
3/ 2
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RI-4. A point charge +Q is brought near a large metal plate and the field lines are as
shown. Does the plate have a non-zero net charge?
A: Yes
B: No
+
Is the magnitude of the charge on the plate greater, less, or the same as the charge Q?
A: greater
B: less
C: same.
Answers: Yes, the plate has a net negative charge, because field lines terminate on it, but
no field lines begin on it.
The plate has a smaller magnitude negative charge than the point charge . There are 8
fields lines coming out of the point charge, but only 5 field lines entering the plate.
RI-5. If the electric field
over a surface be zero

E is non-zero everywhere on a closed surface S, can the flux
 
E  da  0 ?
A: Yes
B: No
z
S

E is constant everywhere on a closed surface S, can you conclude
B: No
 E  da  E  da ? A: Yes
If the electric field
that
S
S
Answers: Yes. A simple example is a constant E-field E  E xˆ . The E-field is non-zero,
 
yet E  da  0 for any closed surface in the field.
z
S
NO!!! In order to "pull the E outside the surface integral" you must first have that
E da everywhere on the surface so that E  da  E da , and then you must have
magnitude of the field = E = constant over the surface.
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RI-6.
A long, straight line of charge has charge per length . Does Gauss's law allow you to
compute the flux
 E  da through the spherical surface of diameter D, centered on the
S
line? A: Yes
B: No

D
Consider now the open hemispherical surface (shaped like a cup) centered on the line as
shown. Does Gauss's law allow you to compute the flux
 E  da through this surface?
S
(Taking the outward normal direction as positive.) A: Yes
B: No

D/2
Answers:
Yes.
 E  da 
S
Qenc
, and Qenc =  D, so
o
 E  da 
S
D
.
o
Yes. Apply Gauss's Law to the closed hemisphere. The surface integral over the flat
portion is zero (since E-field perpendicular to surface there). So we can
D
write  E  da   E  da 
.
2 o
pencup
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RI-7.
The electric field throughout a region of space is given by the formula
E  A y xˆ  B x yˆ , where (x,y) are the coordinates of a point in space, and A, B are
constants. What is E  xˆ ?
A: Ay
B: Bx
C: Ax
D: By
E: None of these.
Answer:
Ay. E  xˆ   A y xˆ  B x yˆ   xˆ  A y xˆ  xˆ  B x yˆ  xˆ  Ay . Since xˆ  yˆ  0 and xˆ  xˆ  1 .
RI-8.
An insulating spherical shell with a uniform positive charge density on its surface is near
a positive point charge. Is the electric field inside the sphere zero?
+
A) E=0 inside
+
+
+
B) E  0 inside
+
+
E=?
+
+
+
C) Not enough info to answer.
+
+
+
+
Answer: E inside is NOT zero. Etotal = Eshell + Epoint Inside, the shell, Eshell =0, but the
field due to the point charge is still present. This is not a metal shell, so there is no
polarization of the charge on the shell and no “screening”.
RI-9. A spherical shell of charge has a non-uniform distribution of charge over its
surface. Is the electric field everywhere within the shell zero?
A: Yes
B: No
E=0?
Answer: No. The E-field inside a uniform, spherical shell of charge is zero. But if the
charge is non-uniform, the E-field will not cancel inside the sphere.
RI-10. Consider a point charge +Q off-center within a spherical metal shell. Does
Gauss's Law allow you to compute the total charge on the inside surface of the shell?
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A: Yes
B: No
Does the total charge on the inside surface of the shell depend on
the total net charge of the whole shell?
A: Yes
B: No
+Q
Consider a dipole within a spherical metal shell.
Is there a non-zero surface charge density  on the
inside surface of the shell?
A: Yes
B: No
+Q -Q
Answers: Yes, you can compute the total charge on the
inside surface. Consider the flux  E  da over a spherical surface within the metal.
S
Since the E-field must be zero everywhere within the conductor, then
 E  da  Qenc / o  0. Since Qenc = 0, there must be a charge -Q spread over the inside
S
surface of the conductor. Since the point charge +Q is off-center, the surface -Q will not
be spread uniformly over the inside surface.
No, the total charge on the inside surface does not depend on the net charge of the shell.
It only depends on the value of the charge within the cavity inside the shell.
Yes, there must be a (non-zero, non-uniform) charge density on the inner surface. Since
the E-field within the conductor must be zero, there must be some charges on the surfaces
of the conductor which create an E-field which cancels the E-field due to the dipole.
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RI-11.Consider two very large, parallel metal plates in static equilibrium. Consider also
the gaussian surface shown. It is proposed that the charge densities on the four surfaces
are as shown. Is this possible?
A: Yes
B: No
+4
-2
+
+
Answer: No. This is not possible. Since E=0 within a conductor, the flux over the
imaginary surface shown must be zero  E  da  0 . Therefore, by Gauss's Law, the
S
enclosed charge must be zero. But if the enclosed charge is zero, the charge densities on
"facing faces" must be equal and opposite.
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©University of Colorado at Boulder