25.11
A Geiger counter has a metal cylinder 2.00 cm in diameter along whose axis is stretched a wire
1.30 ∗ 10−4 cm in diameter. If the potential difference between them is 850V, what is the electric
at the surface of the wire and the cylinder?
λ
êr
where
2π0 r
Z R1
Z R1
E · dr = −
V =−
λ=
E=
R2
R2
V =
Q
L
λ
êr dr
2π0 r
R2
λ
ln(
= 850V
2π0
R1
λ
= 88.17V
2π0
so,
E1 =
25.28
λ 1
= 1.36 ∗ 108 V /m
2π0 R1
and
E2 =
λ 1
= 8.82 ∗ 103 V /m
2π0 R2
In Fig 25-28, what is the net potential at point P due to the four point charges, if V=0 at infinity?
V =
4
X
Vi
i=1
where the potential for a point charge is given by,
Vi =
kq
r
so,
V = V1 + V2 + V3 + V4 =
25.37
k(5.0q) k(−5.0q) k(5.0q) k(−5.0q)
+
+
+
d
d
d
2d
5q
V =
8π0 d
A circular plastic rod of radius R has a positive charge +Q uniformly distributed along one-quarter of
its circumference and a negative charge of -6Q uniformly distributed along the rest of the circumference.
With V=0 at infinity, what is the electric potential at the center, C, of the circle and at the point P, which
is on the central axis of the circle at a distance z from the center?
Let the linear charge density,
λ1 =
Q
π
2
for θ = 0 −
π
2
and
λ2 =
−6Q
3π
2
The electric potential for a continuous charge distribution is found by
Z
kdq
V =
r
1
for θ =
π
− 2π
2
Here,
dq = λdθ
So,
Z
Vc =
Vc =
π
2
0
λ1 dθ
+
R
Z
2π
3π
2
λ2 dθ
R
−5Q
kQ k(−6Q)
+
=
R
R
4π0 R
Since V is independent of the direciton of the radius vector, the equation for Vp is much the same as above.
Z
Vp =
π
2
0
λ1 dθ
+
r
where,
r=
So,
Z
Vp =
0
π
2
Z
2π
3π
2
λ2 dθ
r
p
2
R2 + z 2
λ1 dθ
√
+
2
R2 + z 2
Z
2π
3π
2
λ2 dθ
√
2
R2 + z 2
kQ
k(−6Q)
−5Q
√
Vp = √
+ √
=
2
2
2
2
2
2
R +z
R +z
4π0 2 R2 + z 2
25.39
Figure 25-46 shows a ring of outer radius R and inner radius r = 0.200R; the ring has a uniform surface
charge density σ. With V=0 at infinity, find the expression for the electric potential at point P on
the central axis of the ring, at a distance z = 2.00R from the center of the ring.
The electric potential for a continuous charge distribution is found by
Z
kdq
V =
r
Here, dq = σdA where dA is in polar coordinates and given by dA = r 0 dr 0 dθ
Z
V =
kdq
=
r
Z
0
Z
2π
dθ
1
R
0.200R
kσr 0
√
dr 0
z 2 + r 02
V = 2kπσ(z 2 + r 02 ) 2 |R
0.200R =
25.76
0.113σR
0
What is the excess charge on a conducting sphere of radius, r=0.15 m, if the potential of the sphere
is 1500V and V = 0 at infinity?
For a conducting sphere, the excess charge is uniformly distributed over the surface of the sphere.
The potential of a sphere is given by:
1 Q
V =
4π0 r
So,
Q = 4π0 rV = 4π0 (0.15m)(1500V ) = 2.5 ∗ 10−8 C
2
26.6
You have two flat metal plates, each of area 1.00m2 , with which to construct a parallel-plate
capacitor. If the capacitance of the device is to be 1.00F, what must be the separation between the plates?
The capacitance of a parallel plate capacitor is given by:
C=
Plugging in values,
0 A
d
d = 8.85 ∗ 10−12(1m2 )1F = 8.85 ∗ 10−12 m
This capacitor could not be constructed. The separation distance is too small, on the order of an atom.
26.20
Fig 26-29 shows two capacitors in series; the center section of length b is movable vertically. Show
that the equivalent capacitance of this series combination is independent of the position of the center
0 A
.
section and is given by C = (a−b)
C1
a
b
x
C2
C1 =
0 A
a − (x + b)
and
[C2 =
0 A
x
Since the capacitors are in series, the total capacitance is given by
1
1
1
=
+
Ctot
C1
C2
Ctot =
26.29
1
(a−x−b)+x
0 A
=
0 A
(a − b)
In Figure 26-33, capacitors C1 = 1.0µF and C2 = 3.0µF are each charged to a potential
difference of V = 100V but with opposite polarity as shown. Switched S1 and S2 are now closed.
What is now the potential difference between points a and b? What are now the charges on C1 and C2 ?
Before closing the switches
Q1 = C1 V = 1.0 ∗ 10−6 C ∗ 100V = 1.0 ∗ 10−4 C
and
Q2 = −C2 V = −3.0 ∗ 10−6 C ∗ 100V = −3.0 ∗ 10−4
After closing the switch, the charges will come into equilibrium, so that qtot = qa = −qb
qtot = q1 + q2 = −2.0 ∗ 10−4 C
3
The capacitors are in parallel, so
Ctot = C1 + C2 = 4.0 ∗ 10−6 F
Vab =
qtot
−2.0 ∗ 10−4 C
= −50V = Va − V b
=
Ctot
4.0 ∗ 10−6 F
Now, the charges on the capacitors are
q10 = C1 Vab = 5.0 ∗ 10−5 C
26.30
and
q20 = C2 Vab = 1.5 ∗ 10−4 C
When switch S is thrown to the left in Fig 26-34, the plates of capacitor C1 acquire a potential
difference V0 . Capacitors C2 and C3 are initially uncharged. The switch is now thrown to the
right. What are the final charges q1 , q2 , and q3 on the corresponding capacitors?
The total charge in the circuit is on C1 when the switch is closed. Qtot = C1 V0
After the switch is closed, the total charge redistributes itself over the connection between C1 and C2
so that qtot = C1 V0 = q1 + q2 .
Also, the voltage across C1 is equal to the sum of the voltages across C2 and C3 so that V1 = V2 + V3
giving Cq11 = Cq22 + Cq33 .
The net charges on the lower plate of C2 and upper plate of C3 is 0, so q2 = q3 .
Using these three equations, we solve for the charges:
q2 = q3 =
q1 =
C2 C3
C1 C2 + C1 C3 + C2 C3
C1 C2 + C1 C3
C1 V0
C1 C2 + C1 C3 + C2 C3
4