:`ed aeyigd .orhnd lr divxbhpi` ici lr dcy aygl epilr ~ kdq

:`ed aeyigd .orhnd lr divxbhpi` ici lr dcy aygl epilr
Z
E=
0
L
~ = E î
E
kdq
kλ1 dx
dE = 2 =
r
(P − x)2
L
kλ1 dx
kλ1
kλ1 kλ1
−
=
=
2
(P − x)
P −x 0
P −L
P
jxhvp ,ipyd hend lr lrety gekd z` aygl zpn lr ,
P
mewina dcyd z` mircei epgp`yk eiykr
:divxbhpi` zeyrl
F~ = F î
~
F~ = Eq
kλ1
kλ1
−
λ2 dx
dF = Edq =
x−L
L
Z
3L
F =
2L
kλ1
kλ1
−
x−L
x
λ2 dx = kλ1 λ2 (ln(x − L) − ln(x))3L
2L
F = kλ1 λ2 ln
1
2L · 2L
4
= kλ1 λ2 ln
3L · L
3
Electric dipole
The problem:
1. Prove that if the total charge is zero, then the dipole momenet does not depend on the choice
of the origin.
2. Prove that if the total charge is not zero, then there is an origin such that p~ = 0.
The solution:
1. We know that Q = 0. Let us choose a new set of coordinates such that the new origin is given
by ~b in the original coordinates set:
X
p~ =
qi r~i
(1)
the dipole in the new coordinates is:
X
p~0 =
qi (~
ri − ~b)
X
X
X
p~0 =
qi r~i −
qi~b = p~ − ~b
qi = p~ − ~bQ = p~
(2)
(3)
2. Now Q 6= 0. We need to find a vector that brings us to the point where the dipole is equal to
zero. We choose ~b such that:
p~ − ~bQ = 0
~b = 1 p~
Q
(4)
(5)
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
Energy of a disc and a rod
Submitted by: I.D. 040439358
The problem:
A disc of a radius R is charged uniformly with charge density σ. A rod of a length b is charged
uniformly with charge density λ. The rod is perpendicular to the disc (which is in the x − y plane)
and positioned on the axis of symmetry of the disc. The center of the rod is at z > 2b .
1. Calculate, from the direct integration of the field, the force between the objects.
The solution:
1. The force between the objects
Let ~r1 , ~r2 be the positions of charge elements on the disc and the rod, respectively.
~r1 = (r cos θ, r sin θ, 0)
(1)
~r2 = (0, 0, z)
(2)
~r = ~r2 − ~r1 = (−r cos θ, −r sin θ, z)
(3)
Because of the symmetry of the problem the force is in the z direction only. The electric field due
to the charge element dq on the disc is
dq z
r3
dq = σdA = σrdrdθ
(4)
dEz = k
(5)
The electric field of the disc is
Z R Z 2π
Z R
Z 2π
kσrdrdθz
rdr
Ez =
= kσz
dθ
3
3
0
0
0 (r 2 + z 2 ) 2 0
((r cos θ)2 + (r sin θ)2 + z 2 ) 2 )
Z R
z
rdr
1− √
= 2πkσz
3 = 2πkσ
R2 + z 2
0 (r 2 + z 2 ) 2
(6)
(7)
The force acting on the rod is
F~ =
Z
z+ 2b
z− 2b

Ez ẑλdz = 2πkσλ b +
s
R2 + z −
b
2
s
2
−
R2 + z +
The force acting on the disc is the same but in the opposite direction.
1
b
2
2

 ẑ
(8)
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