Answers  Potential Energy, Potential,
and Capacitors


1.

 The distances from Q_{1} and Q_{2} to
P are r = (y^{2} + a^{2})^{1/2}.
Since V is a scalar quantity, the potentials add algebraically.
V = k(Q_{1}/(y^{2 }+ a^{2})^{1/2
}+^{ }Q_{2}/(y^{2 }+ a^{2})^{1/2})
= k(Q_{1 }+ Q_{2})/(y^{2 }+ a^{2})^{1/2}.
Since a is a constant, V is a function only of y.
 For y^{2} >> a^{2}, V = kQ/y,
where Q = Q_{1 }+ Q_{2} which is the same
as the potential due to a point charge Q located at the
origin.


2.

 The potential V is a scalar quantity. In this case you
may use a minus sign to calculate the potential. For a point
charge Q, the potential at a point P a distance r from P
is V = (9 x 10^{9} Nm^{2}/C^{2})Q/r.
For q_{1}, V_{1} = (9 x 10^{9}
Nm^{2}/C^{2})(9 x 10^{9 }C)/3
m = 27 Nm/C
= 27 J/C = 27 V.
For q_{2}, V_{2 } = (9 x 10^{9
}Nm^{2}/C^{2})(1 x 10^{9}
C)/1 m = 9 V.
V at P = (27  9) V = 18 V.
 The potential at P is really the potential difference
between point P and a very large distance from P. At an
infinite distance, the potential due to q_{1} and
q_{2} is zero (r = a very large number), V_{∞}
= 0. The work done to bring a charge q_{3 }from
a very great distance to P is the increase of the potential
energy of the system. The potential difference V_{P∞}
= the potential energy difference divided by q_{3 }=
(U_{P } U_{∞})/q_{3}.
Work done = q_{3}V_{P∞} = q(V_{P
} V_{∞})= ( 3 x 10^{9} C)(18
J/C  0) = 54 nJ.
 For q_{2} = +1 nC,
V_{1} = (9 x 10^{9} Nm^{2}/C^{2})(9
x 10^{9 }C)/3 m = 27 Nm/C = 27 J/C = 27 V.
V_{2 } = (9 x 10^{9 }Nm^{2}/C^{2})(1
x 10^{9} C)/1 m = 9 V.
V at P = (27 + 9). V = 36 V.
Now the work done = q_{3}(V_{P } V_{∞})
= ( 3 x 10^{9} C){(36 J/C)  0} = 108 nJ.


3.

 In general, the potential due to a point charge Q at a
distance r is kQ/r.
For q_{1} = q_{2} = 200 x 10^{6 }C,
the potential at P is:
V = 9 x 10^{9 }Nm^{2}/C^{2}(2
x 10^{4 }C)/√2 m
= 12.73 x 10^{5} Nm/C
= 12.73 x 10^{5} J/C
= 12.73 x 10^{5} V.
For q_{3} = q_{4} = 100 x 10^{6}
C, the potential at P =
V = 9 x 10^{9 }Nm^{2}/C^{2}(10^{4
}C)/√2 m = 6.36 x 10^{5} V.
Due to all the charges,
V = 2(12.73 + 6.36)10^{5} V
= 12.7 x 10^{5} V.
 Work to bring q_{5} = 20 x 10^{6 }C from
infinity =
q_{5}(V_{P} – V_{∞})
= 20 x 10^{6}C(12.7  0)10^{5 }V = 25.4
J.


4.

 The parallel plate capacitor with its charge and the particle
with the forces acting on it are shown in the figure above.
For the particle to be at rest, the net force acting
on it must equal zero: F_{net } = ma = m(0).
Taking up to be positive, F_{e}  mg = 0
or F_{e }= mg, where the electric force on the particle
with charge q in an electric field E is F_{e} =
Eq. Thus,
Eq = mg or
E = mg/q = (10^{3 }kg)(9.8 m/s^{2})/(10^{6}
C) = 9.8 x 10^{3 }N/C.
The direction of the electric field is up. Remember
the direction of the electric field is that in which a positive
charge is urged.
 V_{ab} = Ed = 9.8 x 10^{3 }N/C (10^{3
}m) = 9.8 J/C = 9.8 V.


5.

This is a problem in conservation of energy. The potential of
a sphere of charge equals the potential of a point charge =
kQ/r. When the alpha particle is a long distance away from the
gold nucleus initially i, the potential energy of the system
U_{i }is 0.
When the alpha particle touches the gold nucleus, r = 7.0 x
10^{15} m = the radius of the gold nucleus. The
potential energy for the "final" position U_{f
} equals:
(9 x 10^{9} Nm^{2}/C^{2})(79)(2)
x (1.6 x 10^{19 }C)^{2}/(7.0 x 10^{15
}m)
= 5.2 x 10^{12 }J.
From conservation of energy,
U_{i} + K_{i} = U_{f} + K_{f}
_{ } 0 + K_{i} =
5.2 x 10^{12} J + 0


6.

U_{i} +
K_{i }= _{ }U_{f}
+ K_{f } or_{ }
U_{i } U_{f } = q(V_{i } V_{f})
= 1/2 mv_{f}^{2}  1/2 mv_{i}^{2
} 1.6 x 10^{19 }C(100 J/C) = 1/2(1.67 x 10^{27}
kg)v_{f}^{2}  0
v_{f} = (2 x 1.6 x 10^{17} J/1.67 x 10^{27}
kg)^{1/2} = 1.38 x 10^{5} m/s


7.

 For the q and + Q system, the potential energy U = kqQ/r
which remains constant if the distance r of separation of
the two charges remains constant.
 The electric force does no work because the force is always
perpendicular to the displacement of the negative charge
when it moves in a circular path around the positive charge.
Thus, there is no potential energy difference between points
along the circular path. The circle is an equipotential
path.


8.

U_{a } U_{b} = q(V_{a
} V_{b})
 For the proton,
U_{a}  U_{b} = 4.8 x 10^{19 }J
= (1.6 x 10^{19} C)(V_{a}  V_{b})
(V_{a}  V_{b}) = V_{ab} =
(4.8 x 10^{19 }J)/(1.6 x 10^{19} C) =
3.0 V.
 For the alpha particle,
U_{a}  U_{b} = q(V_{ab}) = (3.2
x 10^{19 }C)(3.0 V) = 9.6 x 10^{19} J.


9.

(a) V_{A }= 9.0 x 10^{9 }Nm^{2}/C^{2}[80/0.10
 60/0.10]10^{9 }C/m = 1.8 x 10^{3} V
(b)
V_{B} = 9.0 x 10^{9 }Nm^{2}/C^{2}[80/0.16
 60/0.12]10^{9 }C/m = 0
(c) W_{BA}
= q(V_{A}  V_{B}) = (10 x 10^{6}
C)(1.8 x 10^{3} V  0) = 0.018 J


10.

 If the kinetic energy of the object with charge q is to
remain constant, the net force acting on the object must
equal zero. For this to be true,
F_{net }= F + qE
= 0,
where F is the force applied by the outside agent.
The work done in carrying the object from b to a without
increasing its kinetic energy producing a change in potential
energy difference = F ^{. }d = qE ^{.
}d.
 U_{a}  U_{b} = qE ^{. }d
= qEd cos 180^{o} = qEd(1) = qEd
=10^{6}C(10^{4}N/C)(0.10m) = 10^{3
}J.
 qV_{ab} = U_{a}  U_{b }= qEd.
V_{ab} = Ed = (10^{4 }N/C)(0.10 m) = 10^{3
}V.


11.

 W_{BA}
= F ^{. }d, where F is applied by
an external agent and for only an increase in potential
energy F_{net }= F + qE = 0
or F = qE and
W_{BA}
= qE ^{. }s = q(E_{x}s_{x}
+ E_{y}s_{y}) =
10^{6} C[(6)(4) + (8)(4)]10^{4 }Nm/C
= 0.56 J.
For a conservative force, the work done is independent of
the path.
As you go from B to C,  the work done by the electric field
=
10^{6} C(6 x 10^{4} N/C)(4 m) = 0.24
J.
From C to A, work done by the electric field =
10^{6} C(8 x 10^{4} N/C)(4 m) = 0.32
J.
Again the total work done = (0.24 + 0.32)J = 0.56 J.
 U_{A } U_{B }= qE ^{. }s
= 0.56 J.
 qV_{AB} = U_{A}  U_{B}. V_{AB
}= (U_{A } U_{B})/q = 0.56 J/10^{6}
C = 56 x 10^{4 }V.


12.

 V_{B} = kq_{1}/1.0 m + kq_{2}/3.0
m
= 9 x 10^{9} Nm^{2}/C^{2}[(4.0
x 10^{6 }C/1.0 m) +(2.0 x 10^{6} C/3.0
m)]
= 3 x 10^{3}[12 + 2]
J/C = 30 x 10^{3} V
 V_{A }= kq_{1}/3.0 m + kq_{2}/1.0
m
= 9 x 10^{9} Nm^{2}/C^{2}[(4
x 10^{6 }C)/3.0 m) + (2.0 x 10^{6} C/1.0
m)]
= 3 x 10^{3}[4 + 6]
J/C = 6 x 10^{3} V
 Since no work is done to change the kinetic energy of
the particle, the work done to move q_{3} from B
to A =
U_{A}  U_{B} = q_{3}(V_{A
} V_{B}) =
(3.0 x 10^{6} C)[6  (30)] 10^{3}
J/C = 0.108^{} J.
 The electric force is a conservative force so the work
done is independent of the path.


13.

 U = ke^{2}/r
 F_{net } = ma
ke^{2}/r^{2} = mv^{2}/r or
mv^{2} = ke^{2}/r
 E = U + K = ke^{2}/r + 1/2 mv^{2} = ke^{2}/r
+ 1/2 ke^{2}/r = ke^{2}/2r
L = mvr



15.

 When capacitors are wired in parallel, the equivalent
capacitance
C = C_{1} + C_{2} + C_{3}.
But, C_{1} = C_{2} = C_{3} = ε_{0}A/d,
so C = 3 ε_{0}A/d
= ε_{0}A/(d/3).
The plate spacing for the single capacitor must be d/3.
 When capacitors are wired in series, the reciprocal of
the equivalent capacitance 1/C = 1/C_{1 }+ 1/C_{2}
+ 1/C_{3} = 3/C_{1 } and C = C_{1}/3
= ( ε_{0}A/d)/3
= ε_{0}A/3d,
so the plate spacing for the single capacitor must be 3d.


16.

For any capacitance, including C_{1},
C_{1 }= Q/(V_{ab})_{i}, where Q is the
charge on the capacitor when (V_{ab})_{i } is
the potential difference across it. When the two capacitors
are wired in parallel, the potential difference across each
is the same = (V_{ab})_{f } = q_{1}/C_{1}
= q_{2}/C_{2}, where q_{1 }and q_{2}
are the charges on C_{1} and C_{2} with potential
difference (V_{ab})_{f}. Also from conservation
of charge q_{1} + q_{2} = Q or q_{2}
= Q  q_{1} = C_{1} (V_{ab})_{i}
 C_{1 }(V_{ab})_{f} = C_{1}{(V_{ab})_{i
} (V_{ab})_{f}}. But q_{2} also
equals C_{2} (V_{ab})_{f}. Thus C_{1}{(V_{ab})_{i}
 (V_{ab})_{f}} = C_{2}(V_{ab})_{f}
or since C_{1} = 4πε_{0}R_{1
} and C_{2} = 4πε_{0}R_{2,
}4πε_{0}R_{1
}{(V_{ab})_{i}  (V_{ab})_{f}}
= 4πε_{0}R_{2}(V_{ab})_{f
}. Thus R_{2} = R_{1}{(V_{ab})_{i}
 (V_{ab})_{f}}/(V_{ab})_{f}.


17.

The capacitance of a parallel plate capacitor
in a vacuum C = ε_{o}A/d,
where A is the area of the plates and d is the distance between
them.
 When the plate separation is doubled:
 The capacitance C is halved since the capacitance
is inversely proportional to the distance d.
 C = Q/V_{ab} or Q = CV_{ab}. When
C is halved and V_{ab} remains the same (the
battery is still across the capacitor), the charge Q
is halved.
 E = σ/ ε_{o}.
When Q is halved, the charge per unit area σ
is halved and E must be halved. Also E = V_{ab}/d
so when d is doubled E is halved.
 When a dielectric with κ=
2 is inserted:
 The capacitance = κ
e_{o}A/dso the capacitance is doubled.
 Q = CV_{ab}. When the capacitance doubles,
with constant V_{ab}, the charge doubles.
 E = σ/κε_{o}.
When Q doubles, σ doubles,
but σ/κ
remains the same because κ=
2 and E remains the same. Also E = V_{ab}/d
so E is the same because V_{ab} and d remain
the same.


18.

We may view the capacitor as three capacitors in series,
shown in Fig. for #18 above.
1/C_{eq} = 1/C_{1} + 1/C_{2} + 1/C_{3}
= d_{1}/A ε_{0}
+ t/Aκε_{0} +
d_{2}/A ε_{0}.
C_{eq }= A ε_{o}/(d_{1}
+ t/κ + d_{2}).


19.



To find the equivalent capacitance of
Fig. 9 we start with the original circuit.
The capacitors between A’ and C are in parallel so C_{A’C
}= (1 + 2 + 1)µF = 4 µF. The capacitors between
A’ and D are in parallel so C_{A’D} = (6
+ 1 + 1)µF = 8 µF, as shown in Fig. 9 b above.
The capacitors between A’ and B’ are in series.
1/C_{A’B’ }= 1/4 µF + 1/4 µF
= 1/2 µF and C_{A’B} = 2 µF,
as shown in Fig 9c.
The capacitors from A” to B” are in series so
1/C_{A”B’”} = 1/8 µF +
1/8 µF =
1/4 µF and C_{A’B} = 4 µF,
as shown in Fig 9c.
The capacitors between A and B (Fig. 9c) are in parallel and
C_{AB } = 6 µF (Fig. 9d above).
Now work backwards to find the charge.
q_{AB} = C_{AB }V_{AB } = 6 x 10^{6
} (C/v) 12 V = 72 x 10^{6} C = 72 µC.
Up to Fig. 9c, q_{A’B’ } = C_{A’B’
}V_{A’B ’} = 2 x 10^{6 }
(C/V) 12 V = 24 x 10^{6} C = 24 µC. Vq_{A”B”}
= C_{A”B” }V_{A”B”}
= 4 x 10^{6 } (C/V) 12 V = 48 x 10^{6} C
= 48 µC.
Notice q_{A’B’} + q_{A”B”}
= (24 + 48)µC = 72 µC. The two capacitors from A’
to B’ are in series so each has charge = 24µC and
the two capacitors from A” to B” are in series
so each has charge = 48µC. V_{A’C }= V_{DB’
}= 24 x ^{ }10^{ 6 }C/4 x10^{6}
C/V = 6 V.

In Fig. 9e above, from A’ to C, the charge on the two
1µF capacitors is
q_{1}_{µ}_{F} = 1 x
10^{6 }C/V(6V) = 6 x 10^{6} C = 6µC,
and the charge on the 2µF capacitor is
q_{2}_{µ}_{F } =
2 x 10^{6 }C/V(6V) = 12 x 10^{6} C = 12µC.
In Fig. 9e, the charge on the 6µF capacitor is
q_{6µ}_{F } = 6 x 10^{6
}C/V(6V) = 36 x 10^{6} C = 36µC,
_{ }and the charge on the two 1µF capacitors
is
q_{1µ}_{F } = 1 x 10^{6
}C/V(6V) = 6 x 10^{6} C = 6µC.
Check Fig. 9e to see how the charges on the parallel combinations
add up nicely to be equal to the charge of their equivalent
capacitances.


20.

(a) We find the equivalent capacitance by starting with
the parallel combination between D and B.
Capacitors in parallel add. C_{DB} = C_{3}
+ C_{4}= (3.0 + 1.0)µF = 4.0 µF (Fig.
20b).
C_{DB} and C_{2} are in series. The reciprocal
of the equivalent capacitance equals the sum of the reciprocals
of the capacitors in series.
1/C_{A’B’ } = 1/C_{2} + 1/C_{DB}
= 1/12 µF + 1/4.0 µF = (1 + 3)/12 µF or
C_{A’B’ }= 3.0 µF (Fig. 20c).
Then C_{1} and C_{A’B’} are in
parallel. C_{equivalent } = (4.0 + 3.0) µF
= 7.0µF.
(Fig. 20d)
(b) and (c) Now to find the charges and potential differences
we work backwards .
For Fig. 20d,
V_{AB} = 100V = Q_{equivalent}/C_{equivalent}
= Q_{equivalent}/7.0 x 10^{6 }C/V.
Q_{equivalent} = 7.0 x 10^{4} C.
For Fig. 20c,
V_{AB} = 100V = Q_{1}/C_{1} = Q_{1}/4.0
x 10^{6 }C/V.
Q_{1} = 4.0 x 10^{4} C.
Q_{A’B’ } = 100 V(C_{A’B’})
= 100 V (3.0 x 10^{6 }C/V) = 3.0 x 10^{4 }C.
Notice Q_{1} + Q_{A’B’ }= 7.0
x 10^{4 }C.
For Fig. 20b, C_{2} and C_{DB} are in series
and have the same charge.
Q_{2} = Q_{DB} = Q_{A’B’ }
= 3.0 x 10^{4} C.
V_{AD }= Q_{2}/C_{2} = 3.0 x 10^{4}
C/12 x 10^{6 }C/V = 25 V.
V_{DB} = Q_{DB}/C_{DB }= 3.0 x 10^{4}
C/4 x 10^{6 }C/V = 75 V.
Notice V_{AD }+ V_{DB} = 100 V.
Q_{3} = C_{3 }V_{DB} = 3.0 x 10^{6}
C/V (75 V) = 2.25 x 10^{4 }C
Q_{4} = C_{4 }V_{DB} = 1.0 x 10^{6}
C/V (75 V) = 0.75 x 10^{4 }C.
Notice that Q_{3} + Q_{4} = 3.0 x 10^{4
}C = Q_{2}.
(d) Energy stored in a capacitor = U = 1/2 QV = 1/2 Q^{2}/C.
U_{1} = 1/2 Q_{1}^{2}/C_{1 }=
1/2(4.0 x 10^{4} C)^{2}/(4.0 x 10^{6}
C/V) = 2.0 x 10^{2} J.
U_{2} = 1/2 Q_{2}^{2}/C_{2 }=
1/2(3.0 x 10^{4} C)^{2}/(12.0 x 10^{6}
C/V) = 0.375 x 10^{2} J.
U_{3} = 1/2 Q_{3}^{2}/C_{3 }=
1/2(2.25 x 10^{4} C)^{2}/(3.0 x 10^{6}
C/V) = 0.843 x 10^{2} J.
U_{4} = 1/2 Q_{4}^{2}/C_{4 }=
1/2(0.75 x 10^{4} C)^{2}/(1.0 x 10^{6}
C/V) = 0.281 x 10^{2} J.
U_{1} + U_{2} + U_{3} + U_{4}
= 3.5 x 10^{2 }J.
U_{equivalent }= 1/2 (Q_{equivalent})^{2}/C_{equivalent}
= 1/2(7 x 10^{4 }C)^{2}/(7 x 10^{6 }C/V)
= 3.5 x 10^{2} J.

