1.
Three point charges are located along the x axis. The first is located at the point x = -1cm and it is
Q1 = −1µC; the second one is located at x=0 and it is Q2 = 2µC. The third charge is located
at x = 1 cm and it is Q3 = −1µC. Determine the potential and the electric field at the points x = 10cm
and x= -10cm.
V =
3
X
Vi
and
E=
i=1
3
X
Ei r̂
i=1
Where the potential for a point charge is given by,
kqi
ri
Vi =
And the electric field for a point charge is given by,
Ei =
kqi
r̂i
ri2
So,
V = V1 + V2 + V3 =
kq1 kq2
kq3
+
+
x1
x2
x3
For the point at x=10cm.
V =
k(−1 ∗ 10−6 C) k(2 ∗ 10−6 C) k(−1 ∗ 10−6 C)
+
+
= −1818V
.11m
.10m
.09m
At x = -10 cm,
k(−1 ∗ 10−6 C) k(2 ∗ 10−6 C) k(−1 ∗ 10−6 C)
+
+
= −1818V
.09m
.10m
.11m
At x = 10 cm the electric field is given by,
V =
E=
k(2 ∗ 10−6 C)
k(1 ∗ 10−6 C)
k(1 ∗ 10−6 C)
x̂
+
(−x̂)
+
x̂ = 5.5 ∗ 104 N/C
(.11m)2
(.10m)2
(.09m)2
At x = -10 cm.
E=
2.
k(2 ∗ 10−6 C)
k(1 ∗ 10−6 C)
k(1 ∗ 10−6 C)
(−x̂)
+
x̂
+
(−x̂) = −5.5 ∗ 104 N/C
(.09m)2
(.10m)2
(.11m)2
The central lead of a coaxial cable has a diameter, d = 5.0 ∗ 10−5 m; its outside sheath is of diameter
D = 5 ∗ 10−3 m. The space between the central lead and the outside sheath is filled with an insulator of
permeability = 2.70 . The terminals of the battery of EM F = 24V are connected to the
coaxial cable: the positive terminal is connected to the central lead. Determine the direction and magnitude
of the electric field between the two conductors and outside the cable. Determine the energy density of the
electric field at a distance r = 2.0 ∗ 10−3 m from the axis of the cable.
H
Using Gauss’ Law, E · dA =
Inbetween the cylinders,
qenc
and cylindrical symmetry,
E(2πrL) =
λL
so,
1
E=
λ
r̂
2πr
In order to find λ, the charge density, we use the potential of the form
Z
Z
R
d
E · dr = −
∆V = −
r
D
∆V =
λ
r̂ · dr
2πr
D
λ
ln( )
2π
d
So,
∆V
λ
=
2π
ln( D
d)
Plugging into the value for E,
E=
∆V 1
λ 1
=
r̂N/C
2π r
r
ln( D
d)
E=
5.21
r
Outside the two cylinders, λ = 0, so
E=0
The energy density is given by
uE =
For r = 2.0 ∗ 10−3 m,
3.
E 2
2
uE = 8.1 ∗ 10−5 J/m3
An oil drop of mass m = 3.0 ∗ 10−6 kg carrying a charge of Q = 2.0 ∗ 10−7 C is between two horizontal
parallel plates. The plates are a distance of 5.0 cm apart. A variable potential difference is applied to the
plates. Determine the potential difference which keeps the oil drop floating between the plates.
The sum of the forces on the oil drop is equal to zero,
X
F = FE + Fg = 0
Here, FE = QE where the electric field in between two parallel plates is found by V =
E=
R
E · ds So that,
V
d
The gravitational force is given by
Fg = −mg
QV
− mg = 0
d
mgd
= 7.35V
V =
Q
F =
4.
Consider two concentric thin-walled tubes of radii r1 and r2 with (r1¡r2). The inner tube has charge per unit
length λ uniformly distributed on its surface. The outer tube is a conductor with a net charge of zero.
What is the magnitude of the electric field in all regions? What is the charge per unit length on the inner
and outer surfaces of the conductor.?
2
We use Gauss’ Law for this type of problem:
I
E · dA =
qenc
0
(a) For r < r1
qenc = 0
so,
E=0
(b) For r1 < r < r2 drawing a cylindrical gaussian surface so that
E(2πrL) =
λL
0
So,
E=
λ
2π0 r
(c) For r > r2 the conductor has a net charge of zero, so qenc for the gaussian cylinder of r > r2 is
qenc = λL
so,
E=
λ
2π0 r
(d) The electric field inside a conductor = 0. This is created by making qenc = 0 for a gaussian surface inside the
conductor.
so,
qinner = −λL
qinner
= −λ
λinner =
L
(e) To get the charge on the outer layer we use the information that the net conductor charge = 0. So,
qenc = λL + qinner = 0
qconductor = qinner + qouter = 0
giving
λouter =
5.
qouter = −qinner = −(−λL)
qouter
=λ
L
A charge q1 is held fixed at the origin. A second charge q2 initially at rest at r1 is moved
to a new point a distance r2 from the origin, and brought to rest again. Both charges are positive and
r2 > r1 . What is the magnitude of the work done moving the charge? Is energy put into the electric field
What is the magnitude of the average force exerted on q2 in order to move it to its new position?
W = ∆U
where
Ui =
kq1 q2
kq1 q2
and Uf =
r1
r2
W = Uf − Ui = kq1 q2 (
1
1
− )
r1
r2
For r2 > r1 the field gives up energy because
Faverage
6.
W <0
kq1 q2
W
1
1
1
= kq1 q2 ( − )(
=
=
∆x
r1
r2 (r1 − r2 )
r1 r2
Three positive point charges of equal magnitude Q are located at the vertices of an equilateral triangle. Point
P is at the center of the triangle, a distance b from each of the charges. What is the E field at P? What is the
electric potential, V, at P? What is the electrostatic energy, U, of the charge distribution?
3
E=
X
X kqi
Ei =
ri2
r̂i
For three point charges,
E=
kq2
kq3
kq1
r̂1 + 2 r̂2 + 2 r̂3
r12
r2
r3
r1 = r2 = r3 = b
and
q1 = q2 = q3 = Q
but,
r̂1 = ŷ and r̂2 = −cosθx̂ − sinθŷ and r̂3 = cosθx̂ − sinθŷ
√
where
3
2
cosθ =
E=
kQ
(0x̂ + 0ŷ) = 0N/C
b2
The potential is not a vector, so
V = 3(
The energy is
kQ2
kq1 q2
= 3( √ )
r
3b
U =
7.
kQ
)
b
A hollow, spherical shell of inner radius a and outer radius b has a constant charge density. Find the
electric field E for the various regions.
We use Gauss’ Law for this type of problem:
I
E · dA =
qenc
0
(a) For r < a,
qenc = 0
so,
(b) For a < r < b
E(4πr 2 ) =
Z
Z
r
qenc =
r
ρdv =
a
qenc
0
ρr 2 sinθdrdθdφ =
a
E=
E(4πr 2 ) =
Z
b
qenc =
b
ρdv =
a
qenc
0
ρr 2 sinθdrdθdφ =
a
E=
4
πρ(r 3 − a3 )
3
ρ(r 3 − a3 )
3r 2 0
(c) For r > b
Z
E=0
ρ(b3 − a3 )
3r 2 0
4
4
πρ(b3 − a3 )
3
8.
The electric field in the box is parallel to the z-axis everywhere. At the bottom, E = 100V/m. At the top
E =160 V/m. What is the total charge in the box?
I
E · dA =
qenc
0
The surfaces that contribute to the flux have area vectors parallel to the E-field
p
qenc
−(100V /m)(4m)(5m) + (160V /m)cos(30 deg)(4m)( 52 + 2.892 ) =
0
qenc = 1.08 ∗ 10−8 C
Two charges, Q1 and Q2 are fixed on the x-axis; Q1 is at x=a and Q2 is at x=-a.
If the electric field at any point P on the y-axis points in the -x direction, find the ratio of
Q1 to Q2 and the potential at P.
9.
The electric field from a point charge is
E=
kq
r̂
r2
For Q1 , the E field at any point on the y axis is
E=
y
a
kQ1
(−x̂) + p
ŷ
(p
(a2 + y2 )
(a2 + y2 )
(a2 + y2 )
For Q2 , the E field at any point on the y axis is
E=
E=
y
a
kQ2
x̂ + p
ŷ)
(p
2
2
2
2
+y )
(a + y )
(a + y2 )
(a2
(a2
k
(−aQ1 + aQ2 )x̂ + (yQ1 + yQ2 )ŷ)
+ y2 )3/2
In order for Ey = 0,
(yQ1 + yQ2 ) = 0
So,
Q1 = −Q2
V = V1 + V2 = p
10.
so
Q1
= −1
Q2
kQ2
kQ1
−kQ1
kQ1
+p
= p
+p
=0
(a2 + y2 )
(a2 + y2 )
(a2 + y2 )
(a2 + y2 )
Two air gap capacitors are connected in parallel and charged to a certain potential V0 .
The battery is then removed. A dielectric of constant κ is then placed between the plates of C1
Find the final potential across the capacitors and the amount of charge transfered from C1 to C2 .
With the battery attached
Qtot = Ctot V0 = (C1 + C2 )V0
And,
C1 =
5
0 A
d
Now, the battery is removed
C1after =
Ctotafter = κC1 + C2
so,
A
= κC1
d
Vafter =
Qtot
(C1 + C2 )
V0
=
Ctotafter
(κC1 + C2 )
Q1 = C1 V0
Q1after = C1after Vafter =
Qtransferred = Q1after − Q1 = [
11.
κC1 (C1 + C2 )
V0
(κC1 + C2 )
κ(C1 + C2 )
κC2 − C2
− 1]C1 V0 =
C1 V0
(κC1 + C2 )
κC1 + C2
Two point charges of Q = 5C are placed at opposite corners of a square of side d. Two unknown
point charges q1 and q2 are placed at the other corners. It is found that each charge Q
receives zero force. Find q1 and q2 . What is the net force on q1 . What is the total
potential energy in the system.
FQ = Fq1 x̂ + Fq2 ŷ + FQ (cos(45 deg)x̂ + sin(45 deg)ŷ)
In the x direction,
kQq1
0.707kQ2
+
=0
2
d
2d2
q1 = −0.3535Q = 1.76C
FQ =
In the y direction,
kQq2
0.707kQ2
+
=0
d2
2d2
q2 = −0.3535Q = 1.76C
FQ =
kq1 Q
kq1 Q
kq1 q2
k
(−x̂) +
(−ŷ) +
(cos(45 deg)x̂ + sin(45 deg)ŷ) = 2 (−7.7x̂ − 7.7ŷ)
d2
d2
2d2
d
kq1 q2
U=
r
q1 Q q1 Q q2 Q q2 Q
QQ
q1 q2
+
+
+
+√ ]
U = U12 + U13 + U14 + U23 + U34 + U34 = k[ √ +
d
d
d
d
2d
2d
F1 =
U =
12.
−15.31k
k
(2.19 + 4(−8.8) + 17.7) =
d
d
A total charge of Q1 = −5C is uniformly distributed over the volume of spherical shell of inner
radius a = 1 cm and outer radius of b = 2 cm. A point charge of Q2 = 10C is placed at the center
Find the electric field everywhere and find the potential everywhere.
We use Gauss’ Law for this type of problem:
I
E · dA =
qenc
0
(a) For r < a,
qenc = Q2
E(4πr 2 ) =
so,
6
Q2
0
E=
Z
V =−
Z
E · ds = −
b
inf
4.5 ∗ 1010
r̂ · dr −
r2
Z
10C
r̂N/C
4πepsilon0 r 2
a
(−6.4 ∗ 1015 (r −
b
1 ∗ 10−6
)r̂) · dr −
r2
V = 1.125 ∗ 1014 − 8 ∗ 1011 + (9 ∗ 1012 −
(b) For a < r < b
E(4πr 2 ) =
Z
a
r
10C
r̂ · dr
4πepsilon0 r 2
10
9 ∗ 10
)
r
qenc
0
For a uniform distriubtion of charge
3Q1
Q
=
= −1.7 ∗ 105 C/m3
V olume
4π(b3 − a3 )
Z π Z 2π Z r
Z r
4
Q1 (r 3 − a3 )
+ Q2
qenc =
ρdv + Q2 =
ρr 2 sinθdrdθdφ + Q2 = πρ(r 3 − a3 ) + Q2 =
3
(b3 − a3 )
0
0
a
a
ρ=
Q1 (r 3 − a3 )
1
1 ∗ 10−6
15
+
Q
)
=
−6.4
∗
10
(r
−
)r̂N/C
2
(b3 − a3 )
4πr 2 0
r2
Z r
Z
Z b
1 ∗ 10−6
4.5 ∗ 1010
15
r̂
·
dr
−
(−6.4
∗
10
(r
−
)r̂) · dr
V = − E · ds = −
r2
r2
inf
b
E=(
V = 1.125 ∗ 1014 + (6.4 ∗ 1015 (2.5 ∗ 10−5 − 6.4 ∗ 1015 (
(c) For r > b
a3
r2
+ ))
2
r
qenc
0
qenc = Q1 + Q2 = 10C − 5C
E(4πr 2 ) =
4.5 ∗ 1010
r̂N/C
r2
Z
Z r
4.5 ∗ 1010
4.5 ∗ 1010
r̂ · dr =
V = − E · ds = −
2
r
r
inf
E=
13.
Consider a very large planar charge sheet of negligible thickness with a charge density of σ = 3C/m2
Find the electric field everywhere. If a second charge of sheet σ = 2C/m2 is placed parallel
to the first at a distance of 5 cm, find the electric field everywhere.
Use Gauss’ Law with a rectagular gaussian surface:
I
E · dA =
qenc
0
qenc = σA
σ1
σA
so,
E=
= 1.7 ∗ 1011
0
20
σ1
E1 =
= 1.7 ∗ 1011 out of plate
20
σ2
= 1.13 ∗ 1011 out of plate
E2 =
20
E(A + A) =
So, inbetween the plates,
E = E1 − E2 = 0.57 ∗ 1011
7
To the right of both plates
E = E1 + E2 = 2.83 ∗ 1011
To the left of both plates
14.
E = −E1 − E2 = −2.83 ∗ 1011
Two parallel plate capacitors are connected in parallel. The plates have A = 25cm2 and
separated by a distance d = 1mm. One of the capacitors is filled with a dielectric constant of κ = 5
Both capacitors are connected to a V = 5V. Find the total capacitance of the combination. Find the
total charge on the two capacitors. If the dielectric is removed and charge is conserved, what is the new
charge on each capacitor. Find the potential difference across this new combination.
C=
Ctot = C1 + C2 =
A
d
κ0 A 0 A
+
= 1.33 ∗ 10−10 C
d
d
Q = CV
κ0 A
(5V ) = 5.5 ∗ 10−10 C
d
0 A
(5V ) = 1.1 ∗ 10−10 C
Q2 = C2 (5V ) =
d
Q1 = C1 (5V ) =
Q = Q1 + Q2 = 6.6 ∗ 10−10 C
By removing the dielectric material,
C1 = C2 = 2.2 ∗ 10−11F
They are both held at the same voltage, so
Q01 = Q02 = C1 V 0 = C2 V 0
Charge is conserved, so
Q1 + Q2 = Q01 + Q02
Q1 + Q2
= 3.3 ∗ 10−10 C
2
Q0
Q0
V 0 = 1 = 2 = 15V
C1
C2
Q01 = Q02 =
15.
Two point charges are located at the points of coordinates x1 = 0, y1 = 0.5, z1 = 0 and
x2 = 0, y2 = 0, z2 = 1.0. The charges are Q1 = 6µC and Q2 = −6µC. Find the potential at x = 3.0, y = 2.0, z = 1.0
X
kQ1
kQ2
+
r1
r2
p
√
r1 = (∆x)2 + (∆y)2 + (∆z)2 = 9.0 + 2.25 + 1.0 = 3.5m
p
√
r2 = (∆x)2 + (∆y)2 + (∆z)2 = 9.0 + 4.0 + 0 = 3.6m
V =
V =
Vi =
9 ∗ 109 (6 ∗ 10−6C) 9 ∗ 109 (−6 ∗ 10−6C)
+
= 1.74 ∗ 105 V
3.5m
3.6m
8
16.
The plates of a large capacitor are parallel to each other and each have an area A = 4.0cm2 . The
charges on the plates are equal in magnitude, Q = 1 C, and opposite in sign. Find the E field between
the plates. If the separation between the plates is 0.1 cm, find the voltage across the capacitor and
the capacitance of the capacitor.
Using Gauss’ Law,
For one plate,
H
E · dA =
qenc
σ
20
The E fields from the two plates will add together in the center, so
σ
E=
0
E=
σ=
1C
Q
=
= 2500C/m2
A
4.0cm2
E = 2.8 ∗ 1014 N/C
V = E · d = (2.8 ∗ 1014 N/C)(.001m) = 2.8 ∗ 1011 V
0 A
= 3.54 ∗ 10−12 F
C=
d
17.
A very long plastic cylinder of radius, a, is covered with a metal coating of thickness, a. There
is a net uniform volume charge density, ρ, on the plastic cylinder. Find the electric field everywhere.
Find the surface charge density on the metal surfaces at r=a and r=2a.
Using Gauss’ Law,
For r < a,
H
E · dA =
qenc
0
and cylindrical symmetry,
E(2πrL) =
qenc
0
qenc = πr 2 Lρ
So,
E=
rρ
20
For a < r < b, you are inside the metal conductor, so
E=0
For r > 2a, there is no net charge on the metal coating, so
qenc = πa2 Lρ
πa2 Lρ
0
2
a ρ
E=
2r0
In order for the conductor to have not electric field inside, qenc = 0
E(2πrL) =
Qa = πa2 Lρ
σa =
πa2 Lρ
aρ
Q
=
=
Area
2πaL
2
There is no net charge on the conductor, so
σ2a = −σa =
9
−aρ
2