Coulomb law

Coulomb law
Submitted by: I.D. 303689129
The problem:
Two identical rods (length L) are placed on the x̂ axis. The distance between the rods is L. The
rods are charged uniformly, and the total charge on each rod is Q. Find the force on the right rod.
The solution:
The charge density on each rod is
Q
L
dq = λdl
λ =
(1)
(2)
We can sum the Columb forces between the charges on the left rod and the right rod.
X
F~ = k
X
∆qi
X
i
∆qj
j
(~rj − ~ri )
(~rj − ~ri )3
(3)
The index i iterates the charges (positioned at ~ri ) on the left rod. The index j iterates the charges
(positioned at ~rj )on the right rod. The quantity ~rj −~ri gives the distance and the direction between
2 summed charges.
Integrating:
ZL
Fx = k
Z3L
λdl1 λdl2
0
= kλ2
= kλ
2
2L
Z3L
ZL
dl1
0
2L
ZL
dl1
0
= kλ
2
ZL 1
(l2 − l1 )2
(4)
dl2
(l2 − l1 )2
1
l1 − l2
(5)
3L
(6)
2L
1
1
−
dl1
l1 − 3L l1 − 2L
(7)
0
h
L
= kλ2 ln |l1 − 3L|L
0 − ln |l1 − 2L|0
i
(8)
4
3
Q2 4
= k 2 ln
L
3
= kλ2 ln
(9)
(10)
1
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