# SOLUTION Worksheet One: Electrical Potential

```Physics 260 Calculus Physics II: E&M
Grist
SOLUTION
Worksheet One: Electrical Potential
1) When an electron moves from A to B along an electric field line the electric field
does 3.94 1019 J of work on it. What are the electric potential differences:
a) VB  VA , b) VC  VA , c) VC  VB
(a) VB – VA = ΔU/q = –W/(–e) = – (3.94 x 10–19 J)/(–1.60 x 10–19 C) = 2.46 V.
(b) VC – VA = VB – VA = 2.46 V.
(c) VC – VB = 0 (Since C and B are on the same equipotential line).
2) Consider a point charge q  1.0 C , point A at a distance d1  2.0m from q , and
point B at a distance d 2  1.0m .
a). If A and B are diametrically opposite each other, as in the figure (a), what is the
electrical potential difference VA  VB ?
b). What is that electrical potential difference if A and B are located as in figure (b)?
(a) The potential difference is
VA  VB 
q
4 0 rA

1 
 1
 1.0 10 6 C  8.99 109 N  m 2 C 2  


4 0 rB
 2.0 m 1.0 m 
q
 4.5 103 V.

(b) Since V(r) depends only on the magnitude of r , the result is unchanged.
3) A Gaussian sphere or radius 4.0 cm is centered on a ball that has a radius of 1.0
cm and a uniform charge distribution. The total (net) electric flux through the
surface of the Gaussian sphere is 5.60  104 Nm 2 / C . What is the electric potential
12.0cm from the center of the ball?
First using Gauss’ law, solve for the charge, q = εoΦ=+495.8 nC. Next solve for the potential:
V
q
(8.99 109 N  m 2 C2 )(4.958 107 C)

 3.71104 V.
4 0 r
0.120 m
Physics 260 Calculus Physics II: E&M
Grist
4) The electrical potential in the space between two flat parallel plates 1 and 2 is
given (in volts) by V  1500 x 2 , where x (m) is the perpendicular distance from plate
1. At x  1.0cm : What is the magnitude of the electric field and its direction relative
to plate 1?
Use the Del operation that I demonstrated in class. Since in this case the potential is a function of
only one variable this becomes an ordinary derivative:

 dV
E  
 dx
d
ˆ
2 ˆ
2
ˆ
ˆ
ˆ
 i   (1500 x )i  (  3000x)i  (  3000 V/m ) (0.0130 m)i  (39 V/m)i.
dx

(a) The magnitude of the electric field is therefore E = 39 V/m.

(b) The direction of E is  î , or toward plate 1.
5) The electric potential at point in an xy plane is given by
V  ( 2.0V / m 2 )x 2  ( 3.0V / m 2 ) y 2 . What is the electric field at the point 3.0miˆ , 2.0mjˆ ?
Again use the Del operation that I demonstrated in class:
V


(2.0V / m2 ) x 2  3.0V / m2 ) y 2  2(2.0V / m2 ) x;
x
x
V

E y ( x, y)  

(2.0V / m2 ) x 2  3.0V / m2 ) y 2  2(3.0V / m2 ) y .
y
y
E x ( x, y)  
c
c
h
h
We evaluate at x = 3.0 m and y = 2.0 m to obtain

E  ( 12 V/m)iˆ  (12 V/m)jˆ
6) a) What is the electric potential energy of two electrons separated by 2.0 nm?
b) If the separation increases, does the potential; energy increase or decrease
(a) Use the relationship for the potential energy of a pair of point charges with q1 = q2 = –e and r =
2.00 nm:
U k
9
2
2
19
2
q1q2
e 2  8.99 10 N  m C  (1.60 10 C)
k 
 1.15 1019 J.
2.00 109 m
r
r
(b) Since U > 0 and U  r–1 the potential energy U decreases as r increases.
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Nuclear physics

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