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Winter 200315-Phys-202
Solution
Solution
0
0
1 2
012 2 2
15
0
2
0
0
0 20
0 0
0 20
0 09
123
V�ε
q
r
U�ε
q q
r
ε . /
e/e/
.
U�ε
e/ e/
r
e/ e/
r
e/ e/
r
� bx bx x x b
r x
dV�ε
dq
x
V�ε
�dx
x �ε
bxdx
x
b .
�ε
.
� . . �
=1
4
=1
4
= 8 85 10
+2 33
1 32 10
=1
4
(2 3) (2 3)+
(2 3) ( 3)+
(2 3) ( 3)= 0
== 0 = 20 = 20
=
=1
4
=1
4=
1
4=
(0 20)
4
=(20 10 ) (0 20)
4 (8 85 10 )=
1
8 8510 = 36
Prof. R.A. SerotaQuiz 3 Name
Useful formulae and constants:
Potential of a point charge
Potential energy of a pair of point charges
C N m
1. In the quark model of fundamental particles, a proton is composed of three quarks:two �up� quarks, each having charge , and one �down� quark, having charge
. Suppose that the thee quarks are equidistant from one another. Take thedistance to be m and calculate the total potential energy of the three-quark system.
The total consists of all pair-wise terms:
2. A nonuniform linear charge distribution given by , where is a constant, islocated along an axis from to m. If nC/m , what is the electricpotential (relative to potential of zero at in�nity) at the origin?
Here
and
V
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