Oefententamen Elektromagnetisme

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