Electromagnetism Answers
Exercise 1
a
Step 1: Determine the electric field using Gauss’s law
Qenclosed
∮ Eda = ϵ0
For a Gaussian cylinder with radius s and length L the flux becomes:
∮ Eda = E ⋅ 2πsL
So the equition for the electric field is:
Q
E ⋅ 2πsL = ϵ0
If we rewrite for E and write in vector notation we get
Q
E(s) = 2ϵ 0 π sL
s^
Q is just the charge per unit length times its length so:
Q = λL
This means the electric field is
E(s) = λ
2ϵ 0 π s
s^
Step 2: Calculate the potential difference from s to d
s
V = − ∫d E ⋅ dl
In this case this intergral is:
s s 1
V = − ∫d E ⋅ dl = − 2ϵλ0 π ∫d
s
ds = − 2ϵλ0 π ln( sd )
Step 3: Mirror the line charge. Name the line charge above
V+ and the line charge below V_. Calculate the total potential
Vt = V− + V+ =
λ
2ϵ 0 π
ln( sd− ) +
λ
2ϵ 0 π
ln( sd+ ) =
λ
2ϵ 0 π
ln( ss −+ )
Electromagnetism Answers 1
, s+ = y2 + (z − d)2
s− = y2 + (z + d)2 so
y 2 +(z+d)2 2 2
Vt = λ
ln( )= λ
ln( yy 2 +(z+d)
+(z−d)2 )
2ϵ 0 π y 2 +(z−d) 2 4ϵ 0 π
b
The induced surface charge can be calculated with the following formula
σ = −ϵ0 ∂V
∂n
In this case the vector perpendicular to the plane is z. The conducting plane is set
to z = 0. So
σ = −ϵ0 ∂V
∂z ∣z=0
2 2
∂V λ y +(z−d) (y 2 +(z+d)2 )⋅2(z−d)−(y 2 +(z−d)2 )⋅2(z+d)
∂z
= 4ϵ 0 π y 2 +(z+d)2
⋅ (y 2 +(z−d)2 )2
2 2
∂V λ y +d −2d(y 2 +d2 )−(y 2 +d2 )⋅2d
∣
∂z z=0
= 4ϵ 0 π y +d2
2 ⋅ (y 2 +d2 )2
∂V λ −2d−2d λ d
∣
∂z z=0
= 4ϵ 0 π
⋅ (y 2 +d2 )
= ϵ 0 π y 2 +d2
σ = −ϵ0 ∂V ∣
∂z z=0
= − πλ y 2 +d
d
2
Exercise 2
Electromagnetism Answers 2
Exercise 1
a
Step 1: Determine the electric field using Gauss’s law
Qenclosed
∮ Eda = ϵ0
For a Gaussian cylinder with radius s and length L the flux becomes:
∮ Eda = E ⋅ 2πsL
So the equition for the electric field is:
Q
E ⋅ 2πsL = ϵ0
If we rewrite for E and write in vector notation we get
Q
E(s) = 2ϵ 0 π sL
s^
Q is just the charge per unit length times its length so:
Q = λL
This means the electric field is
E(s) = λ
2ϵ 0 π s
s^
Step 2: Calculate the potential difference from s to d
s
V = − ∫d E ⋅ dl
In this case this intergral is:
s s 1
V = − ∫d E ⋅ dl = − 2ϵλ0 π ∫d
s
ds = − 2ϵλ0 π ln( sd )
Step 3: Mirror the line charge. Name the line charge above
V+ and the line charge below V_. Calculate the total potential
Vt = V− + V+ =
λ
2ϵ 0 π
ln( sd− ) +
λ
2ϵ 0 π
ln( sd+ ) =
λ
2ϵ 0 π
ln( ss −+ )
Electromagnetism Answers 1
, s+ = y2 + (z − d)2
s− = y2 + (z + d)2 so
y 2 +(z+d)2 2 2
Vt = λ
ln( )= λ
ln( yy 2 +(z+d)
+(z−d)2 )
2ϵ 0 π y 2 +(z−d) 2 4ϵ 0 π
b
The induced surface charge can be calculated with the following formula
σ = −ϵ0 ∂V
∂n
In this case the vector perpendicular to the plane is z. The conducting plane is set
to z = 0. So
σ = −ϵ0 ∂V
∂z ∣z=0
2 2
∂V λ y +(z−d) (y 2 +(z+d)2 )⋅2(z−d)−(y 2 +(z−d)2 )⋅2(z+d)
∂z
= 4ϵ 0 π y 2 +(z+d)2
⋅ (y 2 +(z−d)2 )2
2 2
∂V λ y +d −2d(y 2 +d2 )−(y 2 +d2 )⋅2d
∣
∂z z=0
= 4ϵ 0 π y +d2
2 ⋅ (y 2 +d2 )2
∂V λ −2d−2d λ d
∣
∂z z=0
= 4ϵ 0 π
⋅ (y 2 +d2 )
= ϵ 0 π y 2 +d2
σ = −ϵ0 ∂V ∣
∂z z=0
= − πλ y 2 +d
d
2
Exercise 2
Electromagnetism Answers 2
