Download Electric field - semicircle

Electric field - semicircle
Submitted by: I.D. 061110185
The problem:
A semicircle of the radius R, 0 < θ < π, is charged with the charge Q.
1. Calculate the electric field at the center if the charge distribution is uniform.
2. Let the charge density be λ = λ0 sin θ. Find λ0 and and the electric field at the center.
The solution:
1) λ - homogeneous
kdq
sin θ
R2
dq = λdl = λRdθ
dEy = −
(1)
(2)
The electric field
~ =
E
Zπ
dEy = −
2kλ
ŷ
R
(3)
0
The charge density is
Z
Z
Q
Q = dq = λRdθ =λRπ ⇒ λ =
Rπ
(4)
So that
~ = − 2kQ ŷ
E
πR2
(5)
2) λ = λ0 sin θ
Z
Q =
λ0 =
Z
dq =
λ0 R sin θdθ = 2λ0 R
(6)
Q
2R
(7)
dq
kλ0
dEy = −k 2 sin θ = − 2 sin2 θdθ
R
R
Z
kλ0 π
Ey =
dEy = − 2
R 2
kQπ
~ = −
E
ŷ
4R2
(8)
(9)
(10)
1
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
Electric dipole
The problem:
An infinite wire is placed on the z-axis and homogeneously charged with linear charge density λ.
An electric dipole p~ = pb
y is positioned at (x, 0, 0).
1. What is the torque acting on the dipole?
2. What is the energy of the dipole in the field of the wire?
3. What is the energy of the wire in the field of the dipole?
4. Find the force acting on the dipole.
The solution:
~ = p~ × E
~
1. We know that the torque acting on a dipole is: N
2kλ
2λ
2λ
r̂ = k 2 ~r = k 2
(xx̂ + y ŷ)
r
r
x + y2
p~ = pŷ
~ = p~ × E
~ = −k 2λ pxẑ
N
x2 + y 2
~ =
E
(1)
(2)
(3)
We can find the torgue at y = 0:
~ (y = 0) = − 2kpλ ẑ
N
x
(4)
2. The energy of the dipole in the field of the wire
U
~ = −pEy = −k
= −~
p·E
x2
2λ
py
+ y2
(5)
(6)
3. The energy of the wire in the field of the dipole is equal to the energy of the dipole in the field
of the wire.
~
4. We can derive the force from the energy by: F~ = −∇U
2x
∂U
F~x = −
= 2kλpy 2
∂x
(x + y 2 )2
(7)
F~x (y = 0) = 0
(8)
x2
y2
−
∂U
F~y = −
= −2kλp 2
∂y
(x + y 2 )2
−2kλp
F~y (y = 0) =
x2
(9)
(10)
~ = ~r × F~ .
And finally we can check ourselves by verifying that N
1