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Quiz #4 Solutions
10 am Section (Section A)
Version
Version
Version
Version
A
B
C
D
1
A
A
D
E
2
C
E
C
D
3
A
E
A
A
4
A
B
B
B
5
C
A
E
A
6
C
E
A
C
11 am Section (Section B)
Version
Version
Version
Version
A
B
C
D
1
E
A
A
C
2
B
B
A
D
3
A
C
C
C
4
D
E
B
C
5
A
D
C
E
6
C
A
E
E
Solutions for version A of each quiz are provided below. Any additional questions
should be directed to:
[email protected]
(Bryan)
OR
[email protected]
(Jason)
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The total work done is path independent, so the easiest thing to do is take each
charge from the 54,000 km triangle to infinity, and then from infinity to the
proper placement around the origin. The work required to take the first charge
from the triangle to infinity is –kq^2(1/r + 1/r) (a 1/r from each of the remaining
charges). Then to take it from infinity to the origin takes no work, since there is
not yet a potential at the origin. The second charge takes –kq^2(1/r) to go to
infinity and +kq^2(1/r) to go to the origin = 0 total. The final charge takes no
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This note was uploaded on 04/06/2008 for the course PHYS 2b taught by Professor Schuller during the Spring '08 term at UCSD.
 Spring '08
 schuller

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