# Exercises in Electrodynamics

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Exercises in Electrodynamics
```Exercises in Electrodynamics
Based on course by Yuri Lyubarsky and Edited By Avry Shirakov
Department of Physics, Ben-Gurion University, Beer-Sheva 84105, Israel
This exercise pool is intended for an undergraduate course in “Electrodynamics 1”.
We start from writing the multipole expansion with some common terms.
Φ(r) =
Z
∞
1 X 1
(r0 )n Pn (cos(θ0 ))ρ(r0 )dV 0
4π0 n=0 rn+1 V 0
So, for n = 0 we get the well known monopole potential
Z
1 Q
1 1
Vmon =
=
ρ(r0 )dV 0
4π0 r
4π0 r V 0
for n = 1 it is the dipole potential
P
Z
1 1
1
r̂i pi
=
r0 cos(θ0 )ρ(r0 )dV 0
Vdip =
4π0 r2
4π0 r2 V 0
and for n = 2 the quadrupole potential
1
=
4π0
P1
Z
r̂i r̂j Qij
1 1
3
1
0 2
2 0
2
=
(r )
cos (θ ) −
ρ(r0 )dV 0
r3
4π0 r3 V 0
2
2
Coming back to the question, by implying the Qij tensor into the equation.
3
X
Z
r̂i r̂j Qij =
(3
0
V
i,j=1
3
X
r̂i ri0
i=1
3
X
r̂j rj0 − (r0 )2
j=1
3
X
r̂i r̂j δij )ρ(r0 )dV 0
i,j=1
now we have to simplify all the expressions that include sums, so the first two equal to
3
X
r̂i ri0 = r̂ · r0 = r0 cos(θ0 )
i=1
and the remaining one equals to
3
X
r̂i r̂j δij =
i,j=1
3
X
r̂j r̂j = r̂ · r̂ = 1
j=1
writing these into the eq we get the following expression,
Z
1 1 1
3r02 cos2 (θ0 ) − r02 ρ(r0 )dV 0
3
4π0 r 2 V 0
=
=
1 1
4π0 r3
Z
1 1
4π0 r3
Z
V0
V0
r02
3
1
cos2 (θ0 ) −
2
2
ρ(r0 )dV 0
r02 P2 (cos(θ0 ))ρ(r0 )dV 0
```
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