Homework 2

Physics 2220 – Bromley Unit 3 HW Solutions Justin – 2013 Spring
The next seven questions pertain to the situation described below.
An infinite line of charge with charge density
λ 1 = 4.6 µC is aligned with the y-axis
cm
as shown.
Figure 1:
1) What is E x (P ), the value of the x-component of the electric field produced by by the line of
charge at point P which is located at (x, y) = (a, 0), where a = 6.7 cm?
∮
∫ ρdV
E · dA =
∂V
ɛ 0
E(2πady) = (λ 1dy)
ɛ 0
E(a) = λ 1
2πɛ 0 a
E(a) = 123582082.521791 N C
2) What is E y (P ), the value of the y-component of the electric field produced by by the line of
charge at point P which is located at (x, y) = (a, 0), where a = 6.7 cm?
The electric field on the x-axis is strictly in the x-direction, |E| = E x , so E y = 0 N C .
3) A cylinder of radius a = 6.7 cm and height
h = 9.2 cm is aligned with its axis along the
y-axis as shown. What is the total flux Φ that
passes through the cylindrical surface? Enter
a positive number if the net flux leaves the
cylinder and a negative number if the net flux
enters the cylinder.
Figure 2:
1

Physics 2220 – Bromley Unit 3 HW Solutions Justin – 2013 Spring
∫ ∫ h
ρdV
2 λ 1 dy
Φ E = = = λ 1h
ɛ 0 − h ɛ
2 0 ɛ 0
4) Another infinite line of charge with charge
density λ 2 = −13.8 µC parallel to the y-axis
cm
is now added at x = 3.35 cm as shown. What
is the new value for E x (P ), the x-component
of the electric field at point P ?
= 4786274.92451412 N · m2
C
Figure 3:
E x (P ) = E 1x (P ) + E 2x (P ) = λ 1
2πɛ 0 a + λ 2
= −617910412.608955 N C
2πɛ 0
(
a −
a
2
) = 1 ( )
λ1
πɛ 0 a 2 + λ 2
5) What is the total flux Φ that now passes through the cylindrical surface? Enter a positive
number if the net flux leaves the cylinder and a negative number if the net flux enters the
cylinder.
Φ =
∫ ρdV
ɛ 0
= 1 ɛ 0
( ∫ L
2
− L 2
λ 1 dy +
∫ L
2
− L 2
λ 2 dy
6) The initial infinite line of charge is now
moved so that it is parallel to the y-axis at
x = −3.35 cm. What is the new value for
E x (P ), the x-component of the electric field
at point P ?
)
= h ɛ 0
(λ 1 + λ 2 ) = −9572549.84902824 N · m2
C
Figure 4:
2

Physics 2220 – Bromley Unit 3 HW Solutions Justin – 2013 Spring
E x (P ) = E 1x (P ) + E 2x (P ) =
λ 1
( ) +
2πɛ 0 a +
a
2
λ 2
2πɛ 0
(
a −
a
2
) = 1 ( )
λ1
πɛ 0 a 3 + λ 2
= −658192847.359847 N C
7) What is the total flux Φ that now passes through the cylindrical surface? Enter a positive
number if the net flux leaves the cylinder and a negative number if the net flux enters the
cylinder.
Φ =
∫ ρdV
ɛ 0
= 1 ɛ 0
( ∫ L
2
− L 2
λ 1 dy +
∫ L
2
− L 2
λ 2 dy
)
= h ɛ 0
(λ 1 + λ 2 ) = −9572549.84902824 N · m2
C
3

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
The next eight questions pertain to the situation described below.
A point charge q 1 = −5.2 µC is located at
the center of a thick conducting shell of inner
radius a = 2.1 cm and outer radius
b = 4.3 cm. The conducting shell has a net
charge of q 2 = 1.1 µC.
Figure 1:
1) What is E x (P ), the value of the x-component of the electric field at point P , located a distance
7.3 cm along the x-axis from q 1 ?
∮
∫ ρdV
E · dA =
∂V
ɛ 0
∮
E dA = 1 ∫ b
)
(q 1 + ρ 2 dV
ɛ 0
E(4πP 2 ) = q 1 + q 2
ɛ 0
a
E x (P ) = q 1 + q 2
4πɛ 0 P 2 = −6914798.710976062 N C
2) What is E y (P ), the value of the y-component of the electric field at point P , located a distance
7.3 cm along the x-axis from q 1 ?
A charge distribution with conformal symmetry creates an equally isotropic electric field.
Thus, along the x-axis, the electric field is strictly in the x-direction, so E y (P ) = 0 N C .
3) What is E x (R), the value of the x-component of the electric field at point R, located a distance
1.05 cm along the y-axis from q 1 ?
Along the y-axis, the electric field is strictly in the y-direction, so at R, E x (R) = 0 N C .
4) What is E y (R), the value of the y-component of the electric field at point R, located a distance
1.05 cm along the y-axis from q 1 ?
∮
∫ ρdV
E · dA =
∂V
E(4πP 2 ) = q 1
ɛ 0
ɛ 0
E y (R) = q 1
4πɛ 0 R 2 = −423902669.3658878 N C
5) What is σ b , the surface charge density at the outer edge of the shell?
1

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
On the inner surface there is a positive induced charge of magnitude −q 1 and a corresponding
negative induced charge on the outer surface of q 1 . The net charge q 2 on the conductor resides
on the outer surface.
By definition, a surface charge density σ is the differential ratio of charge dQ per volume dV .
For a conductor, this ratio must be constant over the entire surface, so in this case σ = Q V .
σ b = Q b
A b
= q 1 + q 2
4πb 2 = −0.0001764562646502896 C m 2
6) What is σ a , the surface charge density at the inner edge of the shell?
The charge density on the inner surface comes from the induced charge only Q a = −q 1 .
σ a = Q a
A a
= −q 1
4πb 2 = 0.0009383284626733058 C m 2
7) For how many values of x: (4.3 cm < x < ∞) is it true that E x = 0?
The electric field is finite for all finite values of x > b, so none .
8) Define E 2 to be equal to the magnitude of the electric field at r = 1.05 cm when the charge on
the outer shell (q 2 ) is equal to 1.1 µC. Define E 0 to be equal to the magnitude of the electric
field at r = 1.05 cm if the charge on the outer shell (q 2 ) were changed to 0. Compare E 2 and
E 0 .
By Gauss’ law, E 2 = E 0 .
2

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
The next eight questions pertain to the situation described below.
An infinite line of charge with linear density
λ 1 = 8.8 µC is positioned along the
m
axis of a thick insulating shell of inner radius
a = 2.0 cm and outer radius b = 4.1 cm.
The insulating shell is uniformly charged with
a volume density of ρ = −670.0 µC
m . 3 Figure 2:
1) What is λ 2 , the linear charge density of the insulating shell?
The relationship between the volume density ρ and linear density λ 2 along the z direction of
the shell is
ρ = dQ
dV = dQ
dAdz = λ 2
dA ,
where A is the cross sectional area π(b 2 − a 2 ). Since ρ is constant, the difference form of this
equation is
Let ∆A = A − 0.
ρ = λ 2
∆A .
λ 2 = πρ(b 2 − a 2 ) = −2.6963324493 µC m
2) What is E x (P ), the value of the x-component of the electric field at point P , located a distance
7.6 cm along the y-axis from the line of charge?
The charge distribution and electric field have cylindrical symmetry (isotropy in r and homogeneity
in z). Thus, the electric field at P will be strictly in the y-direction. Thus,
E x (P ) = 0 N C .
3) What is E y (P ), the value of the y-component of the electric field at point P , located a distance
7.6 cm along the y-axis from the line of charge?
Consider a cylindrical Gaussian surface coaxial with the insulating shell, with length h, and
with radius P .
∮
∫ ρdV
E · dA =
∂V
ɛ 0
E(2πP h) = 1 (∫ h ∫ )
λ 1 dz + ρdV
ɛ 0
0
2πP hE = λ 1h + πρ(b 2 − a 2 )h
ɛ 0
E y (P ) = λ 1 + πρ(b 2 − a 2 )
= 1445605.39029532 N 2πɛ 0 P
C
3

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
4) What is E x (R), the value of the x-component of the electric field at point R, located a distance
1.0 cm along a line that makes an angle of 30 ◦ with the x-axis?
Let θ = 30 ◦ . Consider a cylindrical Gaussian surface coaxial with the insulating shell, with
length h, and with radius R.
∮
∫ ρdV
E · dA =
∂V
E(2πRh) =
∫ h
0
ɛ 0
λ 1 dz
ɛ 0
E(R) = λ 1
2πɛ 0 R
E x (R) = E(R) cos θ = λ 1 cos θ
2πɛ 0 R = 13717835.2196076 N C
5) What is E y (R), the value of the y-component of the electric field at point R, located a distance
1.0 cm along a line that makes an angle of 30 ◦ with the x-axis?
E y (R) = E(R) sin θ = λ 1 sin θ
πɛ 0 R 2
= 7919999.54944 N C
6) For how many values of r: (2.0 cm < r < 4.1 cm) is the magnitude of the electric field equal
to 0?
none
7) If we were to double λ 1 (λ 1 = 17.6 µC ), how would E, the magnitude of the electric field at
m
point P , change?
E would increase by more than a factor of two
8) In order to produce an electric field of zero at some point r > 4.1 cm, how would λ 1 have to
change?
Keep its sign the same and decrease its magnitude
4

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
The next eight questions pertain to the situation described below.
An infinite line of charge with linear density
λ 1 = −5.5 µC is positioned along the axis
m
of a thick conducting shell of inner radius
a = 3.4 cm and outer radius b = 4.5 cm and
infinite length. The conducting shell is uniformly
charged with a linear charge density
λ 2 = 3.2 µC m . Figure 3:
1) What is E x (P ), the electric field at point P , located at (x, y) = (−7.2 cm, 0.0 cm)?
Consider a cylindrical Gaussian surface coaxial with the conducting shell, with length h, and
with radius x = 7.2 cm. Let σ 2 = λ 2
be the surface charge density of the charge uniformly
2πb
distributed on the outer surface of the cylindrical shell.
∮
E · dA = q in
∂V ɛ 0
E(2πxh) = 1 (∫ h ∫ )
λ 1 dz + σdA
ɛ 0
2πxhE = 1 ɛ 0
(λ 1 h + λ 2 h)
0
E(x) = λ 1 + λ 2
2πɛ 0 x = 574999.967288889 N C
2) What is E y (P ), the electric field at point P , located at (x, y) = (−7.2 cm, 0.0 cm)?
Again, due to the radial symmetry (or isotropy in r), the electric field along any line through
the origin is (anti)parallel to that line. Thus E y (P ) = 0 N C .
3) What is E x (R), the electric field at point R, located a distance d R = 1.3 cm from the origin
and making an angle of 30 ◦ with respect to the y-axis as shown?
Consider a cylindrical Gaussian surface coaxial with the conducting shell, with length h, and
5

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
with radius d R . The x-component of the field at this point will be positive.
∮
E · dA = q in
∂V
E(2πd R h) =
ɛ 0
∫ h
0
2πd R hE = λ 1h
ɛ 0
λ 1 dz
ɛ 0
E(d R ) = λ 1
2πɛ 0 d R
E x (d R ) = E(d R ) sin θ = λ 1 sin θ
2πɛ 0 d R
= −3807692.09107692 N C
4) What is E y (R), the electric field at point R, located a distance d R = 1.3 cm from the origin
and making an angle of 30 ◦ with respect to the y-axis as shown?
The y-component of the field at this point is positive.
E y (d R ) = −E(d R ) cos θ = λ 1 sin θ
2πɛ 0 d R
= 6595113.08634979 N C
5) What is λ b , the linear charge density on the outer surface of the conducting shell?
The line of charge along the axis will induce surface charge densities σ in = − λ 1
on the
2πa
inner surface and σ out = λ 1
on the outer surface of the conducting cylindrical shell. The
2πb
net charge density σ 2 on the conducting cylindrical shell resides on the outer surface.
σ b = σ out + σ 2 = λ 1 + λ 2
2πb
Thus, λ b , the total charge density on the outer surface of the shell per length of the shell is
λ b = λ 1 + λ 2 = −2.3 µC m
6) What is λ a , the linear charge density on the inner surface of the conducting shell?
The surface charge density on the inner surface of the shell is due only to the induced charge
density σ in = − λ 1
, or if we only consider charge per length of the cylinder,
2πa
λ a = λ in = −λ 1 = 5.5 µC m
7) The charged conducting shell is now replaced by an uncharged conducting shell of the same
dimensions. What is λ b , the linear charge density on the outer surface of the uncharged
conducting shell?
The net charge on the shell is removed, but the induced charge distributions remain.
λ b = λ out = −5.5 µC m
6

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
µC
8) The conducting shell is now given a new charge (λ 2,new ) such that the electric field at
cm
point P becomes equal to 0. What must be the sign of λ b , the linear charge density on the
outer surface of the charged conducting shell?
The induced charge density on the outer surface is completely determined by the linear charge
along the axis. The net charge on the shell resides on the outer surface. If the total charge
of the system is zero, then the linear charge density at b must be zero, λ b = 0 .
7

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
The next eight questions pertain to the situation described below.
An infinite sheet of charge, oriented perpendicular
to the x-axis, passes through x = 0. It
has a surface charge density σ 1 = −2.2 µC
m 2 .
A thick, infinite conducting slab, also oriented
perpendicular to the x-axis occupies the region
between a = 2.0 cm and b = 4.1 cm. The conducting
slab has a net charge per unit area of
σ 2 = 73.0 µC
m 2 . Figure 4:
1) What is E x (P ), the value of the x-component of the electric field at point P , located a distance
7.6 cm from the infinite sheet of charge?
Consider a cylindrical Gaussian surface with its two circular faces of area A parallel to the
sheet of uniform charge and containing a portion of thereof. The electric field is perpendicular
to the infinitesimal area vector along the round side of the cylinder, so only the circular faces
of the cylinder contribute to the flux out of the cylinder.
∮
E · dA = q in
ɛ 0
EA + 0 + EA = 1 ∫
σ 1 dA
ɛ 0
2EA = σ 1A
ɛ 0
E 1 = σ 1
2ɛ 0
Now consider a similar Gaussian surface containing a cross section of the conducting slab.
The net charge on the slab will be evenly distributed between its two faces.
∮
E · dA = q in
ɛ 0
2EA = σ 2A
ɛ 0
E 2 = σ 2
2ɛ 0
Positive charges originate electric fields and negative charges terminate electric fields, so in
the region containing P , the sign on E 1 is negative and the sign on E 2 is positive in the
x-direction. For any point to the right of the slab, and in particular point P , the electric
field is
E = −|E 1 | + |E 2 | = σ 1 + σ 2
2ɛ 0
= 3998108.0966039775 N C .
8

Physics 2220 – Bromley Unit 4 HW Solutions Justin – 2013 Spring
2) What is E y (P ), the value of the y-component of the electric field at point P , located a distance
7.6 cm from the infinite sheet of charge?
The electric fields due to each charge distribution σ 1 , σ 2 are only in the ±x-directions, so
their sum, the net field also only has x-components. Thus, E y (P ) = 0 N C .
3) What is E x (R), the value of the x-component of the electric field at point R, located a distance
1.0 cm from the infinite sheet of charge?
In this region the sign on E 1 is still negative while the sign on E 2 is now negative.
E = −|E 1 | − |E 2 | = σ 1 − σ 2
2ɛ 0
= −4246578.091308179 N C .
4) What is E y (R), the value of the y-component of the electric field at point R, located a distance
1.0 cm from the infinite sheet of charge?
The fields only have nonzero x-components, so E y (P ) = 0 N C .
5) What is σ b , the charge per unit area on the surface of the slab located at x = 4.1 cm?
The net charge on the conducting slab distributes evenly between the two surfaces, and the
plane of charge at x = 0 will induce charge densities − σ 1
2 at a and σ 1
at b (such that the
2
net electric field inside the slab is zero). Thus, the net charge on the surface at x = b is
σ b = σ 2
2 + σ 1
2
= 35.4
µC
m 2
6) What is E x , the value of the x-component of the electric field at a point on the x-axis located
at x = 2.84 cm?
This point is inside the conducting slab, so the field is zero. In particular, the x-component
of the field E x = 0 N C .
7) What is σ a , the charge per unit area on the surface of the slab located at x = 2.0 cm?
σ a = σ 2
2 − σ 1
2
= 37.6
µC
m 2
8) Where along the x-axis is the magnitude of the electric field equal to zero?
The fields produced by each charge distribution are uniform and homogeneous, thus the net
field can only be zero inside the conductor, so the answer is none of these regions .
9