. .

[ ]

∫ ∫

�

�

�

�

� �

��

� �

�� �

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