Gauss' Law Homework Solutions
1. A %Þ! cm-square in the B-C plane sits in a uniform electric field E œ a#Þ! i  $Þ! j  &Þ! kb NÎC.
Find the electric flux through the square.
Since the square is in the B-C plane, only electric field in the (perpendicular) D -direction contributes
to the flux. Therefore:
FI œ I¼ E œ a&Þ! NÎCb ‚ ˆ"' cm# ‰ œ )Þ! ‚ "!$ N † m# ÎC
(Alternatively, you can use FI œ E † A, where A œ "' k cm# is the area vector.)
2. A thundercloud produces a vertical electric field of magnitude #)Þ! kNÎC at ground level. You hold
a ##Þ! cm ‚ #)Þ! cm sheet of paper horizontally below the cloud.
(a) What is the electric flux through the sheet?
FI œ IE œ ˆ#)Þ! ‚ "!$ NÎC‰ˆ!Þ!'"' m# ‰ œ "Þ(# ‚ "!$ N † m# ÎC
(b) What would the flux be if you tilt the sheet of paper by $!°?
FI œ IE cos ) œ ˆ"Þ(# ‚ "!$ N † m# ÎC‰a!Þ)''!b œ "Þ%* ‚ "!$ N † m# ÎC
(c) What would the flux be if you hold the sheet of paper vertically?
Since the electric field would be parallel to the paper, the flux would be zero .
3. A cylindrical metal can has a height of #( cm and a radius of "" cm. The electric field is directed
outward along the entire surface of the can (including the top and bottom), with a uniform magnitude
of %Þ! ‚ "!& NÎC. How much charge does the can contain?
The area of the side of the can is #1<2 œ #1a!Þ"" mba!Þ#( mb œ !Þ")'' m# , the area of the top is
1<# œ 1a!Þ"" mb# œ !Þ!$)! m# , and the area of the bottom is the same as the top. Thus the total
area of the surface is
ˆ!Þ")'' m# ‰  ˆ!Þ!$)! m# ‰  ˆ!Þ!$)! m# ‰ œ !Þ#'#' m# .
Therefore, the total flux out of the can is
FI œ IE œ ˆ%Þ! ‚ "!& NÎC‰ˆ!Þ#'#' m# ‰ œ "Þ!&!% ‚ "!% N † m# ÎC.
By Gauss' Law, the charge inside is %! FI , which is *Þ$ ‚ "!( C .
(The charge is positive since the electric field is directed outward.)
4. A 'Þ!-nC point charge is located at the center of a cube of side length #Þ! m. What is the electric
flux through each of the faces of the cube?
By Gauss' Law, the total flux coming out of the cube is
FI œ UÎ%! œ ˆ'Þ! ‚ "!* C‰Îˆ)Þ)& ‚ "!"# C# ÎN † m# ‰ œ '() N † m# ÎC.
By symmetry, this flux must be evenly split between the six faces. Therefore, the flux through each
face is "' a'() N † m# ÎG b œ ""! N † m# ÎC .
5. A spherical ball of plasma has a radius of !Þ&! m. The electric field on the surface of the ball is
measured to be )Þ* ‚ "!$ NÎC and points radially inward toward the center. What is the average
volume charge density of the plasma?
By Gauss' Law, the charge inside is
U œ %! FI œ %! IE œ %! I%1<# œ ˆ)Þ)& ‚ "!"# ‰ˆ)Þ* ‚ "!$ ‰%1a!Þ&!b# œ #Þ%( ‚ "!( C.
Thus the volume charge density is
U
3œ
œ
volume
U
a#Þ%( ‚ "!( Cb
(
œ
CÎm$ .
$ œ %Þ( ‚ "!
%
$
a%Î$b1a!Þ&!b
$ 1<
(The charge is negative since the electric field is pointing inwards.)
6. In the air over a particular region at an altitude of '!! m above the ground, the electric field is
"#! NÎC directed downward. At '&! m above the ground, the electric field is "!! N/C downward.
What is the average volume charge density in the layer of air between these two elevations?
Consider a large box of length P, width [ , and height &! m, sitting so that its base is at an altitude
of '!! m, and its top is at an altitude of '&! m. Then the flux out of the bottom of the box is
Fout œ IE œ a"#! NÎCbP[
and the flux into the top is
Fin œ IE œ a"!! NÎCbP[ .
By Gauss' Law the total charge inside the box is
Uinside œ %! aFout  Fin b œ ˆ"Þ(( ‚ "!"! CÎm# ‰P[ .
The volume of the box is a&! mbP[ , so the average volume charge density inside the box is
3 œ
U
a"Þ(( ‚ "!"! CÎm# bP[
œ
œ $Þ& ‚ "!"# CÎm$ .
volume
a&! mbP[
(The charge is positive since there is more flux out than flux in.)
7. Charge is placed on a metallic sphere which is surrounded by air. If the radius of the sphere
is "& cm, how much charge can be placed on the sphere before the air near the sphere suffers electric
breakdown? The critical electric field strength that leads to breakdown in air is $Þ! ‚ "!' NÎC.
If we place a charge U on the sphere, the electric field outside the sphere is governed by Coulomb's
law
I œ
" U
%1%! <#
In particular, the electric field immediately outside the sphere is
I œ
"
U
œ ˆ%Þ! ‚ "!"" NÎC# ‰U
%1%! a!Þ"& mb#
The air will break down when I œ $Þ! ‚ "!' NÎC, so the maximum possible U is
$Þ! ‚ "!' NÎC
œ (Þ& ‚ "!' C
%Þ! ‚ "!"" NÎC#
8. A long copper wire with a radius of "Þ! mm carries a uniform surface charge density of &Þ! µCÎm# .
(a) Find the total charge in a "Þ!-meter-long section of the wire.
A "Þ! m long section of wire has a surface area of
#1<2 œ #1ˆ"Þ! ‚ "!$ m‰a"Þ! mb œ 'Þ#) ‚ "!$ m#
and therefore has a charge of a&Þ! ‚ "!' CÎm# ba'Þ#) ‚ "!$ m# b œ $Þ" ‚ "!) C
(b) Find the magnitude of the electric field at a distance of "& cm from the wire.
As determined in part (a), the linear charge density of the wire is - œ $Þ" ‚ "!) CÎm. Therefore,
the electric field at a distance of "& cm is
I œ
" " $Þ" ‚ "!) CÎm
œ
œ $Þ) ‚ "!$ NÎC .
#1%! <
#1%!
!Þ"& m
9. An insulating sphere with a radius of #! cm carries a uniform volume charge density of "Þ& µCÎm$ .
Find the magnitude of the electric field at a point inside the sphere that lies )Þ! cm from the center.
A point )Þ! cm from the center will only feel a force from the charge within a radius of )Þ! cm. (The
charge between < œ )Þ! cm and < œ #! cm will not affect it) The total amount of charge within a
radius of )Þ! cm is
%
U œ 3Z œ ˆ"Þ& ‚ "!' CÎm$ ‰ 1a!Þ!)! mb$ œ $Þ#"( ‚ "!* C
$
This charge has the shape of a sphere, so the electric field is determined by Coulomb's law:
I œ
" U
" $Þ#"( ‚ "!* C
œ
œ %Þ& ‚ "!$ NÎC
%1%! <#
%1%! a!Þ!)! mb#
10. A square metal plate with a thickness of "Þ& cm has no net charge and is placed in a region of
uniform electric field )Þ! ‚ "!% NÎC directed perpendicularly to the plate. Find the resulting surface
charge density on each face of the plate.
Since the plate is conductive, the electric field will induce positive and negative charges on opposite
sides of the plate so that the total electric field inside the plate is zero. Suppose that the surface
charge density on each side of the plate is 5. Then the total electric field will be
5
5
Itotal œ ˆ)Þ! ‚ "!% NÎC‰ 

#% !
#% !
Setting this equal to zero and solving for 5 gives
5 œ %! ˆ)Þ! ‚ "!% NÎC‰ œ (Þ" ‚ "!( CÎm#