Download Recitation #4 Solution

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Physics 212 - Spring 2000
Recitation Activity 4: Gauss' Law
NAME: Solution
Instructor: Redwing DATE: January 25, 2000
This activity is based on the following concepts:
 Electric FLUX through a surface is defined in the following way:
 Divide the surface into small patches
 Multiply the area of a patch by the component of the electric field in the direction
perpendicular to the patch
 Add the contributions from all patches on the surface -- that total sum is the flux!
 In practice, we make the patches infinitesimally small, and hence the sum is
replaced by an INTEGRAL.
 The electric flux  through a Gaussian surface is proportional to the net number


of electric field lines passing through that surface:    E  dA
S



Gauss' Law is a FUNDAMENTAL law of nature that says:
 The electric flux through any CLOSED surface = (TOTAL charge enclosed)
divided by 0
Q
 We write this mathematically as:  E  dA 
0
S
The electric field inside a metal (at equilibrium) is ZERO!
Some math that you MUST review:
 What is the surface area of a sphere of radius R?
 What is the volume of a sphere of radius R?
 What is the surface area of a cylinder of length L and radius R?
 What is the volume of a cylinder of length L and radius R?
Exercise 1: Understanding Electric Flux
The figures below show different surfaces in the presence of different electric fields.
Determine the electric flux in each case, briefly showing your reasoning.
(1)
(a) Rectangle of area A; uniform electric field E pointing
parallel to area A.




FLUX =    E  dA   E (cos 90)dA  E (0) A  0 , since the
S
S
normal to the surface is perpendicular to E.
600
(1)
(b) Rectangle of area A; uniform electric field E
making an angle of 600 with the NORMAL to the
area A.

 

FLUX =    E  dA   E (cos 60)dA  EA cos 60
S
S
(1)
(c ) Spherical surface of radius R; electric field
pointing radially outwards from the center of the
sphere. Magnitude of the electric field everywhere on
the surface of sphere is E.




FLUX =    E  dA   E (cos 0)dA  EA  E ( 4R 2 )
S
Exercise 2: Applying Gauss' Law
The figure below shows a point charge +Q surrounded
by an UNCHARGED spherical metallic shell of inner
radius R1 and an outer radius R2. Your goal is to use
Gauss' Law to find out the magnitude of the electric field
at the points A and C that are located at distances RA and
RC from the center of the sphere as shown in the figure.
You will also use Gauss' Law to determine the charge
induced on the inner surface of the spherical shell.
Remember to use the spherical symmetry of the problem
in your arguments.
(1)
B
A
C
R1
+Q
R2
(a) To figure out the field at A, construct a spherical Gaussian surface of radius RA.
What is the electric flux through this surface?

 

2
FLUX =    E  dA   E (cos 0)dA  EA  E ( 4R A )
S
(1)
S
S
Using Gauss' Law, what is the electric field at A?


  Q
2
   E  dA  E ( 4R A )
 E  dA 
0
S
E (4RA ) 
2
Q
0
S
E
Q
2
40 RA
(b) Now, develop an argument to figure out the charge on the inside and outside surfaces
of the spherical shell. Use the following concepts to do this: (i) Gauss’ Law (ii) the
knowledge that E = 0 inside a metal under conditions of equilibrium (iii) the spherical
symmetry of the problem.
Inside shell, r < R1, E 
Q
40 r 2
Inside metal, E = 0



  Q
E

d
A

(
0
)

d
A
 0 , so Q = 0.
,
but
E

d
A




S
(1)
)
(1)
)
0
S
S
Since there is +Q inside shell, there must be –Q at r = R1 to make the total charge inside
for R1 < r < R2 equal to zero, i.e. –Q on the inside surface of the shell.
Since there was initially no net charge on the conductor, there must be +Q amount of
charge some place to cancel the –Q on the inside surface. The only place left is r = R2.
So, +Q amount of charge on the outside surface of the shell.
(c) Finally, figure out the electric field at C by constructing a spherical Gaussian surface
of radius RC.
What is the electric flux through this surface?
(1)




2
FLUX =    E  dA   E (cos 0)dA  EA  E ( 4RC )
S
S
)
What is the TOTAL charge enclosed by this surface?
(1)
)
(1)
)
+Q –Q +Q = +Q
Using Gauss' Law, determine the electric field at C.
  Q
E
  dA 
S
0
 
2
   E  dA  E ( 4RC )
S
Q
Q
2
E (4RC )   E 
2
0
40 RC
Scoring Guidelines:
100
90
75
60
40
0
All correct
8.5 – 9.9
3 - 8.4
0 - 3 correct, but shows effort
Turned in paper, but no real effort
No paper