Download 18.1 Review PP

Chapter 18
Capacitance
and Potential
1. What is the electric
potential energy
between a proton and
an electron separated
by a distance of
-11
5.3 x 10 m?
PE = k q1 q2 / d
EPE = (9 x 109)(1.6 x 10-19)(-1.6 x 10-19)/(5.3 x 10-11 )
PE = -4.35 x
-18
10
J
The charges are opposite so the
energy is negative.
2. Two positive point
charges are separated by
a distance of 4 cm. If they
are moved to a distance
of 12 cm apart, by what
factor does the electric
potential energy between
them change?
PE = k q1 q2 / d
Three times as far away, 1/3 the
potential energy.
This is NOT an inverse-square
relationship.
3. A particle of charge
-15
3 x 10 C moves 5 m in
the direction of a uniform
electric field of strength
4000 N/C. What is the
change in electrical
potential energy?
PEE = qEd
PEE = (3 x
-15
10 )4000(5)
PEE = -6 x
-11
10
J
(The positive charge is moving in the
direction it wants to go, so it is a
decrease in potential energy.)
4. If you move the
particle in problem 3
back to its starting
position, what is the
change in electrical
potential energy for this
second move?
It is being moved in the
direction it doesn’t want
to go, so it is an
increase in potential
energy.
+6 x
-11
10
J
5. What is the
strength of the
electric field in
problem 3?
Duh, 4000 N/C.
I wanted you to calculate
it using Ed = V, but I
messed up!
6. Two point charges,
+2.7 μC and +3.8 μC,
are 0.55 m apart. What
is the electrical
potential energy of this
system of two charges?
PE = k q1 q2 / d
EPE = (9 x 109)(+2.7 x 10-6)(+3.8 x 10-6)/(0.55 m)
PE = 0.168 J
The charges are the same so the
energy is positive.
7. What would be the
electrical potential
energy of the system in
problem 6 if both
charges were negative
instead of positive?
The same, PE =
0.168 J, two negative
charges would still
produce a positive
potential energy by
PE = k q1 q2 / d.
-19
10 )
8. A proton (q = 1.6 x
moves 3 meters along an
electric field of 500 N/C.
What is the potential
difference between the
proton’s beginning and final
location?
Ed = V
500 x 3 = V
1500 V = V
Moving in the direction of the field
is a decrease of potential, so
-1500 V = V.
9. A uniform electric field of
1000 N/C is directed
downward. If the potential at
10 m above the ground is
15,000 V, what is the
potential at ground level? At
3 m above the ground?
Ed = V, so the potential difference
between 10 m and 0 m is 1000 x
10 = 10000 V. Since the field is
directed downward, the potential
must be 10000 V lower at the
ground: 15000 -10000 = 5000 V. At
3 m above the ground the potential
must be 1000 x 3 higher: 3000 +
5000 = 8000 V.
10. What is the
electric potential
2 m from a point
charge of 22 μC?
V = k q/ d
V = (9 x
9
10 )(22
x
-6
10 )
/2
V = 99000 V
Potential is a scalar, so there is
no direction associated with this
answer.
11. Two point charges are
separated by 3 meters. If
the charges are 25 μC
and 13 μC, what is the
potential at the midway
point between the two
charges?
V = k q/ d
V1 = (9 x 109)(25 x 10-6) /1.5
V2 = (9 x 109)(13 x 10-6) /1.5
V1 = 150000 V
V2 = 78000 V
V = 150000 + 78000 =
228000 V
12. Two point charges are
separated by 4 meters. If
the charges are +36 μC
and -24 μC, what is the
potential at the midway
point between the two
charges?
V = k q/ d
V+ = (9 x 109)(36 x 10-6) /2
9
-6
V- = (9 x 10 )(-24 x 10 ) /2
V+ = 162000 V
V- = -108000 V
V = 162000 + -108000 =
54000 V
13. At what distance
from a point charge
of 16 μC is the
5
potential 2 x 10 V?
V = k q/ d
2 x 105 = (9 x 109)(16 x 10-6) / d
d = 0.72 m
14. At what
distance from a
point charge of
16 μC is the
potential zero V?
The potential
is always zero
at infinity.