CT27-10. A uniform, infinite plane of negative charge creates a uniform E-field of
magnitude E perpendicular to the plane and pointing toward the plane as shown. An
imaginary gaussian surface in the shape of a right cylinder is shown. (This shape is
sometimes called a "pillbox".) The flux through surface is..
E
E
cap area A
radius r
L
A) –EA
E) None of these
C) (–2A+Lr2) E
B) 2EA
D) Lr2E
Answer: None of these, because the flux is -2EA (negative).
  The flux through the side of

the cylinder is zero because on the sides, E  da and E  da  0 The flux through each
end cap is -EA, since on the end caps E is anti-parallel to da and E  da  0 .
da
E
E
CT27-11.
da
The non-zero electric field everywhere on a closed surface is constant:
r
r
E = constant (meaning vector E is everywhere constant in magnitude and
direction). Is the following calculation correct?
r r
E
òÑ ×da = E òÑda = EA
A) Definitely correct
B) Definitely incorrect
C) Possibly correct, possibly incorrect – depends on details of the surface and E.
Answer: Definitely incorrect. It is only true that
r
r
r
r r
òÑE ×da = E òÑda
IF,
everywhere on the surface, E || da , and E = E = constant . On any closed surface,
r
the various da ' s on the surface point in various directions (always along the local
outward normal). So if the E-field always points in the same direction (say along the +xaxis) then there is no way the E-field can everywhere be parallel to the
r
da ' s .
CT27-12. A negative point charge with charge -Q sits in the interior of a spherical metal
shell. The conducting metal shell has no net charge. What is the total charge on the inner
surface of the shell?
inner surface
outer surface
-
gaussian
surface
A) –Q
B) +Q
C) +2Q
D) zero
E) Some other answer.
Hint: Consider the gaussian surface
shown. Answer: +Q
Answer: Apply Gauss's law to the gaussian surface shown,
 E  da  Q
in
/  o  0 , since
E=0 everywhere on the surface, inside the metal. So there must be a +Q on the inside
surface, canceling the -Q point charge, making Qenclosed=0.
What is the total charge on the exterior surface of the shell?
A) –Q
B) +Q
C) +2Q
D) zero
E) Some other answer.
Answer: -Q Since the net charge on the metal shell is zero, if there is +Q on the inside
surface, there must be -Q on the outside surface.
The field lines looks like this:
E=0 (inside)
CT27-13.
A sphere of radius R has a total charge +Q spread uniformly throughout its
volume. The charge density is  
Q
. (The sphere must be an
(4 / 3) R 3
insulator, because it can't be a metal. Why not?) We are going to compute
the electric field magnitude E within the sphere.
What is the charge enclosed by the centered small sphere of radius r?
A) Q(4/3)r3
R3
D) Q 3
r
r2
B) Q 2
R
r3
C) Q 3
R
E) None of these.
If the electric field at distance r from the center of the sphere has magnitude E, what is
the flux
z
 
E  da through the small sphere, radius r?
2
A) E(R )
D) E(4/3)r2
B) E(r2)
E) None of these.
C) E(4R2)
Within the sphere, the electric field magnitude E is proportional to
A) r
B) r2
C) r3
D) None of these, E=constant within the sphere.
E) None of these, E=0 within the sphere.
R
r
Answers:
r3
Part 1: The charge in the little sphere is Q 3 . The charge inside the small sphere is Q
R
times the fraction of the total volume in the small sphere = Q
(4 / 3) r 3
r3
.

Q
(4 / 3) r 3
R3
Part 2: None of these! The flux thru the small sphere is
E(4r2) = (field at small sphere) times (area of small sphere).
Part 3: Inside the sphere, E  r. Gauss's Law says
 E  da  Q
in
/ o
r3 1
 E(4r )  Q 3 
R o
2
Solving for E gives E 
F
Q I
G
H4R  J
Kr,
3
for r  R .
o
It is easy to show, from Gauss's Law, that, outside the big sphere, r > R, the E-field is just
that due to a point charge Q,
E
Q
4  o R 2
E
~r
~ 1/r2
r
R