x
z
y
R
EE140 MT1 Spring 2021 Dr. Ray Kwok

1. A thin wire, carrying a positive and uniform linear charge density , is bent into a quarter-circle with a radius
R, as shown in the figure. (a) Find the electric field vector at a point (0, 0, z). (b) Find the electric potential at
the same location using = −∫ ⃗ ∙ ℓ⃗ .

⃗ ()
2
〈−, −, 〉

=

3
〈−, −, 〉
= √2 + 2 = √2 + 2 + 2 = constant
= −
2
3
∫

2
0
= −
2
3
=

=

3
∫

2
0
=

23

⃗ =

3
〈−,−,

2
〉

= − ∫⃗ ∙ = −

2
∫

[2 + 2]3/2

∞

∞

= 2 + 2, = 2
∫

[2 + 2]3/2

∞
=
1
2
∫−3/2 = −−
1
2 = −[
1
√2 + 2
]
∞

= −
1
√2 + 2

=

2√2 + 2


Check with direct integral:
=


=


, =


∫
/2
0
=


∙

2
=

2
=

2√2 + 2






r
x
z
dq (x,y,0)
dE
y
R

(0,0,z)
Why ∞ to z? Can it be -∞ to z?
Why dz? Can it be dx? dy?
by symmetry
EE140 MT1 Spring 2021 Dr. Ray Kwok

2. The electric field in front of a thin charged circular disc at a distance z away
from the center is given by (from lecture, homework and quiz problems):
⃗ = ̂

2
[1 −

√2 + 2
].
Use this result to find the magnitude of the electric field at the tip of a right-
circular cone has a base radius R, height h, and a uniform charge density  as
shown in the figure. z
h
R
Quiz: Ring to disc
Quiz: Q on solid cone
EE140 MT1 Spring 2021 Dr. Ray Kwok

2. The electric field in front of a thin charged circular disc at a distance z away
from the center is given by (from lecture, homework and quiz problems):
⃗ = ̂

2
[1 −√2 + 2].
Use this result to find the magnitude of the electric field at the tip of a right-
circular cone has a base radius R, height h, and a uniform charge density  as
shown in the figure. z



2 ↔ 2, → , ℎ > > 0
() =
()
2
[1 −√2 + 2]=ℎ=2∫[1 −√2 + (ℎ )2]
ℎ0=2∫ [1 −ℎ√ℎ2 + 2]
ℎ0=ℎ2[1 −ℎ√ℎ2 + 2]
h
R
z
r
h
R
EE140 MT1 Spring 2021 Dr. Ray Kwok

3. An infinitely long thin wire carrying a uniform charge density , is placed in front of the grounded conductor
at (d, s, 0) and parallel to the z-axis as shown. (a) Find the electric field at any point (x, y, z) in the first quadrant.
You do not need to simplify your answer, as long as all the variables are CLEARLY defined. (b) Find the
surface charge density induced on the xz-plane (i.e., at y = 0).




Grounded
conductor
line
charge s
d
y
x
air
EE140 MT1 Spring 2021 Dr. Ray Kwok

3. An infinitely long thin wire carrying a uniform charge density , is placed in front of the grounded conductor
at (s, d, 0) and parallel to the z-axis as shown. (a) Find the electric field at any point (x, y, z) in the first quadrant.
You do not need to simplify your answer, as long as all the variables are CLEARLY defined. (b) Find the
surface charge density induced on the xz-plane (i.e., at y = 0).

E field for an infinite wire (Gauss’s Law):
=

2


⃗ 1 =

21
2
〈 − , − , 0〉
⃗ 1 =

2[( − )2 + ( − )2]
〈 − , − , 0〉
Similarly,
⃗ 2 =
−
2[( + )2 + ( − )2]
〈 + , − , 0〉
⃗ 3 =

2[( + )2 + ( + )2]
〈 + , + , 0〉
⃗ 4 =
−
2[( − )2 + ( + )2]
〈 − , + , 0〉
⃗ (, , ) = ⃗ 1 + ⃗ 2 + ⃗ 3 + ⃗ 4

As y = 0, the only non-zero component is the y-component (normal to conductor).
( = 0) =
(−)
2[( − )2 + 2]
−
(−)
2[( + )2 + 2]
+
()
2[( + )2 + 2]
−
()
2[( − )2 + 2]

Simplify, call it the normal component, and remember D = E (in air)

= −


{
1
[( − )2 + 2]
−
1
[( + )2 + 2]
}
Knowing D = 0 inside the conductor,
( − ) = = −


{
1
[( − )2 + 2]
−
1
[( + )2 + 2]
}
which is always negative (x > 0), as expected.







Grounded
conductor
line
charge s
d
y
x
(x,y,z)
r1
L1+
L3+
L2-
L4-
air
EE140 MT1 Spring 2021 Dr. Ray Kwok

4. A long dielectric cylinder of radius “a”, r1 = 2, is carrying a volume charge density of  = r. It is completely
enclosed by another cylinder of outer radius “b”, r2 = 4, and has a uniform charge density = -5 C/m3. Find E(r)
in all regions, including the air region outside. No plot is needed.


























a
b
r2 = 4
r1 = 2
EE140 MT1 Spring 2021 Dr. Ray Kwok

4. A long dielectric cylinder of radius “a”, r1 = 2, is carrying a volume charge density of  = r. It is completely
enclosed by another cylinder of outer radius “b”, r2 = 4, and has a uniform charge density = -5 C/m3. Find E(r)
in all regions, including the air region outside. No plot is needed.

For r < a,
∮⃗ ∙ =
(2) = ∫ = ∫(3)(2)

0
= 23
= 2 = 2
⃗ ( < ) =
2
2
̂
For b > r > a,
(2) = 23 − 5[(2 − 2)]
=
3

−
5(2 − 2)
2
= 4
⃗ ( > > ) =
23 − 5(2 − 2)
8
̂

For r > b,
(2) = 23 − 5[(2 − 2)]
=
3

−
5(2 − 2)
2
=
⃗ ( > ) =
23 − 5(2 − 2)
2
̂








a
b
r2 = 4
r1 = 2






















































































































































































































































































































































































































































































































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