SOLUTIONS: PROBLEM SET 2
ELECTRIC POTENTIAL AND ELECTRIC POTENIAL ENERGY
PART A: CONCEPTUAL QUESTIONS
A. Electrons are free to move in a conductor. If there was a potential difference between two points, then an
electric field must exist. Charges will be pushed in this electric field, and will redistributed themselves until
the electric field is zero. Once there is no electric field, charges do not gain or lose potential energy if they
move from point to point. All points are at the same potential.
B. It might be frightening to think you are at 5000V, but remember: potential is energy per charge. So if there’s not
a lot of charge, there’s not really a lot of energy involved. Each coulomb of charge may have a potential
energy of 5000J relative to the ground, but if you only have 10µC on your body, that’s only 0.05J! The story
is different if you have a source of charge that is significantly larger. Like a lightning bolt of HydroQuebec…
Consider an analogous situation with gravity: the gravitational potential at a point
500 m above the
ground is 5000J/kg (taking the ground as 0), but it makes a big
difference if a grain of chalk dust is held over
your head at that point of it a car is!
C. a)
D. a)
E. b)
F. b)
G. b)
H.
15V
20V
P
20V
E = 1000 N/C = 1000 V/m
25V
I. E (although D is pretty close)
J. A
K. 80 V
1
L. C
M. 2x10-5J
N. & O. (diagram)
P. 8x10-5J
Q. 4x10-5J
PART B: NUMERICAL QUESTIONS
QUESTION 1
The potential energy between charges is given by: U =
kq1q2
r
We can find the distance between the charges
kq1q2 k (2.2 ×10−6 )(3.48 ×10−6 )
r=
=
U
0.4
= 0.172m
b) The potential energy is related to kinetic energy: ΔK + ΔU = 0
K f − Ki = U i − U f
1 2
mv f − Ki = −( U f − 0.4)
2
solve for the speed:
v=
2(0.4)
1.5 ×10−5
v = 231m / s
2
QUESTION 2
a) The electric potential energy of several charges is the sum of contribution of individual charge.
⎛ q q q q q q ⎞
U = k ⎜ 1 2 + 1 3 + 2 3 ⎟
r13
r23 ⎠
⎝ r12
⎛ (2 ×10−6 )(−4 ×10−6 ) (2 ×10−6 )(3 ×10−6 ) ( −4 ×10−6 )(3 ×10 −6 ) ⎞
U = k ⎜
+
+
⎟
0.04
0.05
0.03
⎝
⎠
U = −4.32 J
b) The electric potential of several charges is the algebraic sum of the individual contribution.
⎡ q
q
q ⎤
V p = k ⎢ 1 + 2 + 3 ⎥
⎣⎢ r1 p r2 p r3 p ⎥⎦
⎡ 2 × 10−6 −4 × 10−6 3 × 10−6 ⎤
= k ⎢
+
+
⎥
0.05
0.04 ⎦
⎣ 0.03
V p = 5.55 ×105V
c) The charge q at point P would have a potential energy
U = qV p
ΔK + ΔU = 0
K f − K i = −( U f − U i )
1 2
mv = qV p
2
2qV p
⇒v=
m
=
2(3 ×10−10 )(5.55 ×105 )
4.2 ×10−20
v = 8.90 ×107 m / s
3
QUESTION 3
The kinetic energy and potential energy are related to the electric potential.
ΔK + ΔU = 0 = qVBA
K f = Ki + ΔK = Ki + (−ΔU )
1
K f = (2 ×10−4 )52 + [−(5 ×10−6 )(800 − 200)]
2
K f = 2.5 ×10−3 + 3 ×10−3 = 5.5 ×10−3 J
1 2
mv f = 5.5 × 10 −3 J
2
2K f
⇒ vf =
m
Kf =
2(5.5 × 10−3 )
2 × 10−4
=
v f = 7.42m / s
QUESTION 4
The kinetic energy and potential energy are related to the electric potential
ΔU = −ΔK
qV3,0 = −( K f − Ki )
⇒ V3,0 =
−( K f − K i )
q
⎡ (1.5 ×10−2 ) − (5 ×10−2 ) ⎤
= − ⎢
⎥
−6.5 ×10−7
⎣
⎦
V3,0 = −5.38 ×104V
V at origin is 5.38x104 V higher than V at 3 cm.
A positive charge going from 0 to 3 would lose potential energy and gain kinetic energy. The reverse happened
for this negatively charged particle.
4
QUESTION 5
The electric field and electric potential are related together.
VAB = VA − VB =
B
∫A
r r
E ⋅ dl and that leads to V = Ed for uniform electric field only.
The electric field of a plate is given in terms of the surface charge density σ
σ
1×10−3
E=
=
= 5.65 ×107 N / C
−12
2ε o 2(8.85 ×10 )
VAB = VA − VB =
B
∫A
r r
Edl = E
.5
∫.2 dx
VAB = 0.3E
VAB = 1.70 ×107V
b) The work is also the change in potential energy.
WB→ A = U A − U B = VAB q
WB→ A = 1.7 ×107 (1×10−9 )
WB→ A = 1.70 ×10−2 J
5
QUESTION 6
Consider point P, x meter to the left of q1:
Vp =
k q1
k q2
+
=0
x
( x + 2)
q1
q
=− 2
x
( x + 2)
→
(q1 + q2 ) x = −2q1
⇒x=−
2(2 µC )
−1µC
x = 4m
to the left of q1
Consider a different point P, in between q1 and q2
Vp =
k q1
k q2
+
=0
x
(2 − x)
q1
q
=− 2
x
(2 − x)
→
(−q1 + q2 ) x = −2q1
⇒x=
−2(2µC )
−5µC
x = 0.8m
to the right of q1
What about point P to the right of q2
Vp =
k q1 k q2
+
=0
2+ x
x
→
q1
q
=− 2
2+ x
x
6
(q1 + q2 ) x = −2(q2 )
⇒x=
−2(3µC )
−1µC
x = −6m
to the right of q 2 !?
Answers: 4 m to the left of q1 ad 0.8m to the right of q1
b)
VA =
k q1
k q2
+
=0
y
4 + y2
→
q1
q2
=−
y
4 + y2
q12 (4 + y 2 ) = q22 y 2
y 2 (q12 − q22 ) = −4q12
y=
−4(2 µC ) 2
(2 µC ) 2 − ( −3µC ) 2
y = ±1.79 m
N.B. The equipotential V= 0 is a continuous surface (I’m not sure what its actual shape is); what we’ve done is find
the 4 points where it cuts through the axes.
7
c) No.
First of all, by inspection, it is clear that E cannot be 0 at any of the points except, maybe (-4, 0)m. and a quick
calculation shows that it is not equal to 0 here as well. More generally, the value for V anywhere depend on the
arbitrary choice of a reference zero and does not depend upon the E at the location in question. For example, a
table top might be chosen as a level of 0 gravitational potential energy; that doesn’t mean that the gravitational field
is 0 at the level.
QUESTION 7
How close it will get?
The energy conservation states that:
( Etot ) far = ( Etot ) near
Vq = U + K =
kQq
r
kQ
V
k (79 ×1.6 ×10−19 )
=
3 ×106
r=
r = 3.79 ×10−14 m
The force:
kqQ
r2
k (2 ×1.6 ×10−19 )(79 ×1.6 ×10−19 )
=
(3.79 ×10−14 )2
F=
F = 25.3N
WOW!
The acceleration:
a=
=
F
m
25.3
4 × 1.67 × 10−27
a = 3.79 × 1027 m / s 2
8
QUESTION 8.
a) Recall your physics NYA and energy stored in a string. The energy conservation states that
( Etot ) A = ( Etot ) B
1
0 = kx 2 + (−VAB q)
2
VAB =
kx 2
2(0.05) 2
=
2q 2(6 × 10−5 )
VAB = 41.7V
b) The electric field is related to the electric potential if the electric field is uniform by:
E=
VAB 41.7
=
Δx 0.05
E = 834V / m
c) for a uniform electric field:
V = Ed = 834V / m × 0.2 m
V = 167V
d) Let’s draw a free body diagram(!!)
ΣF = FE − FS = 0
qE = k Δx
→ Δx =
→ FE = FS
Eq 834V / m ⋅ 6 ×10−5 C
=
= 0.025m
k
2N / m
Answer: 2.5 cm. Note that this is the midpoint of the block’s
oscillation.
9