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



8.

 dA = 2 πr dr
 dq = σ dA = 2πσrdr
 dV = kdq/r = 2πσkrdr/(z^{2}
+ r^{2})^{1/2}



9.

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


10.


In the field of these
charged wires, draw the circular equipotential traces
passing through C and D. The electric field
due to the wire through A or B = λ/2πε_{o}r,
where r is the distance from the wire to the field points.
The direction of the field is radially away from the wire.

First consider the wire passing though A:
In the field of both wires, the potential difference is twice
as large, or
λ/2πε_{o}(2)
= λ/πε


11.

 When capacitors are wired in parallel, the equivalent
capacitance
C = C_{1} + C_{2} + C_{3}
But C_{1} = C_{2} = C_{3} = ε_{o}A/d
so C = 3 ε_{o}A/d
= ε_{o}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_{o}A/d)/3
= ε_{o}A/3d
so the plate spacing for the single capacitor must be 3d.


12.

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_{o}R_{1
} and
C_{2} = 4πε_{o}R_{2,
}4πε_{o}R_{1
}{(V_{ab})_{i}  (V_{ab})_{f}}
= 4πε_{o}R_{2}(V_{ab})_{f},
so
R_{2} = R_{1}{(V_{ab})_{i} 
(V_{ab})_{f}}/(V_{ab})_{f}.


13.

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 = κε_{o}A/d
so 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.


14.

The electric field in the air gaps is  E_{o}j_{
} =  σ/ ε_{o}
j=  Q/A ε_{o}
j.
The electric field in the dielectric is  Ej =  Q/Aκε_{o
}j.
V_{ab } =  _{a}∫^{b} E^{
. }dyj = (Q/A ε_{o})(d_{1}
+ t/κ + d_{2}).
C = Q/V_{ab} = A ε_{o}/(d_{1}
+ t/κ + d_{2}).
Alternatively, we may view the capacitor as three capacitors
in series, shown in the lower portion of the figure above.
For series 1/C_{eq} = 1/C_{1} + 1/C_{2}
+ 1/C_{3} = d_{1}/A ε_{o}
+ t/Aκε_{o} + d_{2}/A ε_{o}.
C_{eq
}= A ε_{o}/(d_{1}
+ t/κ + d_{2}).


15.

 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. 4b ).
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. 5c).
Then C_{1} and C_{A’B’}
are in parallel.
C_{equivalent } = (4.0 + 3.0) µF = 7.0 µF.
 Now to find the charges and potential differences we
work backwards.
For Fig. 4d,
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. 4c,
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. 4b, 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}.
 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.



17.

From Gauss' theorem, we found that the
electric field a distance r from the axis of a cylinder equals
2k λ/r, where λis the charge per unit length. For a cylinder of total
charge Q and length L, λ=
Q/L and E = 2kQ/rL. In this problem, we have coaxial cylinders
of radius a and radius b with b > a. Using Gauss's
theorem, the electric field between the cylinders still equals
2kQ/rL because a Gaussian surface there does not enclose the
charge on the outer cylinder.
 By definition,
 C = Q/V_{ab} = Q/[(2kQ/L) ln b/a] = L/[2k (ln
b/a)].


18.

To move the plates a distance ds, takes
dW = F ds, where F is the force we are seeking.
 The energy stored in the electric field is U = (1/2)Q^{2}/C.
For a parallel plate capacitor of area A and separation
s,
C = ε_{o}A/s
and
U = Q^{2}s/2 ε_{o}A,
where Q, ε_{o},
and A are constants.
 The work dW would appear as an increase dU in the potential
energy:
dU = (Q^{2}/2 ε_{o}A)ds
Equating dW and dU, we find
F ds = (Q^{2}/2 ε_{o}A)ds
and
F = (Q^{2}/2 ε_{o}A).


19.

 q = (charge/volume)(volume) = ρ(4 πr^{3}/3)
 q’ = ρ(4 πr^{2}
dr)
 dU = kqq’/r = kρ^{2}(4 π)^{2}r^{5}
dr/3r
 U = kρ^{2}(4 π)^{2}/3
_{o}∫^{R} r^{4} dr = kρ^{2}(4 π)^{2}
R^{5}/15 = 3k(4πR^{3}ρ/3)^{2}/5R
= 3kQ^{2}/5R, where the total charge of the sphere
of radius R is Q = (4 πR^{3}ρ/3).


20.

3kQ^{2}/5r_{o} = mc^{2}.
r_{o} = 3kQ^{2}/5mc^{2
} = 3(9.0 x 10^{9 }Nm^{2}/C^{2})(1.6
x 10^{19 }C)^{2}/5(9.1 x 10^{31}
kg)(3.0 x 10^{8 }m/s)^{2
} = 1.7 x 10^{15} m.


21.

 From conservation of angular momentum,
L_{A
}= L_{B
} r_{A}mv_{A} sin 90^{o }=
r_{B}mv_{B} sin 90^{o
} v_{B}/v_{A }= r_{A}/r_{B
}= 6 x 10^{5}/3 x 10^{5} = 2
 From conservation of energy,
_{ }U_{A }+
K_{A }=
U_{B } +
_{ } K_{B
} kqq’/r_{A }+ 1/2 mv_{A}^{2
} = kqq’/r_{B }+ 1/2 mv_{B}^{2
} kqq’/r_{A }+ kqq’/r_{B}^{
}=_{ }1/2 mv_{B}^{2 } 1/2
mv_{A}^{2
} kqq’(1/r_{B } 1/r_{A}) = ^{
}m/2(v_{B}^{2 } v_{A}^{2})
= m/2(4v_{A}^{2 } v_{A}^{2})
= 3mv_{A}^{2}/2
9 x 10^{9} Nm^{2}/C^{2}(10^{6
}C)(10^{6} C)(10^{5}/m)[1/3 1/6]
= 3(10^{10} kg)v_{A}^{2}/2
v_{A }= 10^{6 }m/s
v_{B} = 2v_{A} = 2 x 10^{6} m/s


22.


 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





25.

In general,
In all cases, E is constant over the Gaussian surface
and the angle between
E and dA is zero.
Thus,
.
 For a < r < b, the dashed Gaussian surface
is inside a conductor so the electric field is zero. The
charge enclosed must be zero or
0 = q_{R } + q_{a} and
q_{R }=  q_{a} = 5 µC.
 For 0 < r < R, the Gaussian surface
is again inside a conductor and E = 0.
For R < r < a, the charge enclosed is
q_{a}.
E4 πr^{2 }= q_{R}/ ε_{o}.
E = 5 x 10^{6}C/4πε_{o}r^{2}
= k(5 x 10^{6}C/r^{2}).
The minus sign means the electric field for R <
r < a is radially inward.
For a < r < b, as we said before, E
= 0.
For r > b, the charge enclosed = q_{R
}+ q_{a} + q_{b} = q_{b} since
q_{R }=  q_{a}.
E4 πr^{2 }=q_{b}/ ε_{o.}
And E = q_{b}/4πε_{o}r^{2}
= kq_{b}/r^{2}.

For r >b, V(r) = kq_{b}/r.
For a < r < b, V(r) = kq_{b}/b.
For R < r < a, V(r) = kq_{b}/b
 kq_{a}(1/r  1/a).
For r < a, V(r) = kq_{b}/b  kq_{a}(1/R
 1/a).
Plots of E and V as a function of r are below in Fig. for
25.


26.

 E(r) = λ/2πε_{o
}
 ln ∞ ≠ 0. With an infinite wire you
can't get away with choosing V = 0
at infinity.
 E = dV/dr = d{( λ/2πε_{o})(ln
a – ln r)/dr = λ/2πε_{o}r!


27.

E =  dV/dx
or
E =  slope of V vs
x
For –0.4 m < x < 0.3 m,
 Slope =  (150 – 0)V/{(0.3 –(0.4)}m
= +1500 V/m.
For –0.3 m < x < 0.1 m,
 Slope = {150 – (150)}V/{(0.1 –(0.3)}m
= 1500 V/m.
For –0.1 m < x < 0.1 m,
slope = 0.
For 0.1 m < x < 0.3 m,
 slope = (150 – 150)V/(0.3 –0.1)}m
= +1500 V/m.
For 0.3 m < x < 0.4 m,
 slope = {0 – (150)}V/(0.4 –0.3)}m
= 1500 V/m.
If you are so inclined, you can work backwards.
V – 0 = ∫ E dx or
V =  area under E vs
x.
For example,  the area from x = 0.40 m
to x = 0.3 m is 150 V.
V(0.3m) = 0 –150 V =  150 V. Try it
for a few other points.

