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The electric field
Submitted by: I.D. 040460479
The problem:
An isolated electrical wire is charged uniformly with charge q and bent into a circular shape (radius
R) with a small hole b R (where b is the arc length).
What is the electrical field in the middle of the circle?
The solution:
The simple solution is to use superposition.
The electric field in the middle of a complete ring is, of course, zero.
Now we’ll sum up the field of a complete ring with the field of a very small wire of the same size
and shape as the hole but with a negative charge.
~ = kλb
x̂
Because b R the wire can be taken as a negative point charge and, therefore, the field is E
R2
q
(when we take the hole to be on the X axis and λ = 2πR ).
It is also possible to calculate the field directly. Taking ~r = (0, 0, 0) and ~r0 = (R cos θ, R sin θ, 0) we
have
kdq
(−R cos θ, −R sin θ, 0)
R3
dq = λRdθ
Z −α
kλ
kλRdθ0
~
(−R cos θ0 , −R sin θ0 , 0) =
(2 sin α, 0, 0)
E =
3
R
R
α
~ =
dE
(1)
(2)
(3)
where ±α are the angles of the edges of the bent wire (or, the upper and lower limits of the hole).
Since b R we can approximate tan α ' sin α = b/2
R .
Substituting into the expression for the field we obtain
~ = kλb x̂
E
R2
(4)
1
BGU Physics Dept. Physics 2 for physics students
Cylinder and dipole
1. Volume charge density of sphere- ρ = Q/πR2 L. The electric field on the z axis is
2π
Z
Ez =
Z
LZ R
ρ(b − z)rdrdzdφ
=
(r2 + (b − z)2 )3/2
p
p
2Q 2 + (b − L)2 −
2 + b2
L
+
R
R
πR2 L
0
=
0
(1)
0
(2)
2. The volume charge density is ρ = ρ0 cos φ. The straightforward way to solve this problem is
to write down the expression of the field and expand it in a taylor series. However, we know
that far away from the cylinder, b L, R, this charge distribution looks like a dipole in the
x̂ direction. So there is no point in going through all the derivation again. We just need the
dipole moment p~ and then field is
~ = p~ · r̂ r̂ − p~
E
r3
r3
(3)
The dipole moment is
Z
Z
ρ(~r)~rd3~r = ρ0 cos φ(r cos φ, r sin φ, z)rdrdφdz =
Z L Z Z 2π
πR3 Lρ0
x̂
= x̂ρ0
cos2 φr2 drdφdz =
3
0
0R 0
p~ =
(4)
(5)
We want the field at ~r = bẑ, so
3
~ ≈ − πR Lρ0 x̂
E
3b3
(6)
1