Download Coulomb law

Coulomb law
Submitted by: I.D. 066101528
The problem:
Two small spheres of identical masses and charges are hung on two strings of a length L. Show that
for small angles the horizontal distance between the masses is:
2
1/3
q L
x=
(1)
2πε0 mg
The solution:
We write two force equations one for the vertical axis (where θ is the angle of the string relatively
to the vertical):
T cos θ = mg
(2)
and another one for the horizontal axis (x is the distance between the masses):
T sin θ =
kq 2
x2
(3)
Divide the equations and use the following formulas:
tan θ =
k =
sin θ =
Kq 2
mgx2
1
4πε0
x/2
' tan θ
L
(4)
(5)
(6)
where in the last equation we used the the small angles. Then
q2
4πε0 mgx2
1/3
2
q L
x =
2πε0 mg
x
2L
=
(7)
(8)
1
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
Submitted by: I.D. 039609631
The problem:
Two identical particles, with the charge +q, are held in place at a distance d from each other. A
particle charged −Q with the mass m is placed between the two, in the middle. The particle (+Q)
is slightly diverted upwardas, perpendicularly to the straight line connecting the two particles (+q),
and let go. Show that harmonic oscillations occur, maintaining:
T2 =
ε0 mπ 3 d3
Qq
(1)
The solution:
We define the axis on which the possitive charges lay as x. y axis is perpendicular to x, positive
direction set upwards. Negative charge Q is set on axis origin. ~r1 and ~r2 are position vectors for
positively charged particles.
Coulomb Law :
kqqi (~r − ~ri )
ΣF~ =
(2)
k~r − ~ri k3
For our problem we get :
kq(−Q)(−~r1 ) kq(−Q)(−~r2 )
ΣF~ =
+
(3)
k~r1 k3
k~r2 k3
Where :
d
x̂ − (∆y) ŷ
(4)
~r1 =
2
d
x̂ − (∆y) ŷ
(5)
~r2 = −
2
Therefore :
kQq[ d2 x̂ − (∆y)ŷ] kQq[ − d2 x̂ − (∆y)ŷ]
−2kQq∆y
~
ΣF =
+
=
ŷ
(6)
2
3
3
3
2
2
[ d2 + ∆y 2 ] 2
[ d2 + ∆y 2 ] 2
[ d2 + ∆y 2 ] 2
simplifying, and substituting k with ε0 (k ≡
ΣF~ =
1
4πε0 )
we get :
−4Qq∆y
2 3 ŷ
2∆y
3
πε0 d [1 + d
]2
(7)
Extracting taylor series to first order, using the relation:
1
3
x + O(x2 )
3 = 1 −
2
(1 + x) 2
We get :
−4Qq∆y
−4Qq∆y
ΣF~ =
ŷ or mÿ =
πε0 d3
πε0 d3
⇒ ω2 =
4Qq
mπε0 d3
where
T2 =
4π 2
ω2
(8)
(9)
(10)
Finally
T2 =
ε0 mπ 3 d3
Qq
(11)
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 field
Submitted by: I.D. 060645009
The problem:
• Calculate directly the electric field flux through an infinite plane a distance d from a point
charge.
• Explain why your result does not depend on d.
The solution:
Let the plane to be at the x − y plane, so that any point on it is ~r = (r cos(ϕ), r sin(ϕ), 0) and the
point charge is at ~r0 = (0, 0, d). By the Coloumb law
kdq(~r − ~r0 )
|~r − ~r0 |3
~ =
dE
(1)
In our case
~r − ~r0 = (r cos(ϕ), r sin(ϕ), −d)
0
2
(2)
2 1/2
|~r − ~r | = (r + d )
(3)
So that
~ =
E
(r2
kq
(r cos(ϕ), r sin(ϕ), −d)
+ d2 )3/2
The flus is
I
Z
~ =
~ · dA
E
Ez dA
Z 2π
Z
=
dϕ
0
Z
= 2πkqd
0
(4)
(5)
∞
0
∞
rdr
kqd
3
(r2 + d2 ) 2
rdr
3 = 2πkq
(r2 + d2 ) 2
(6)
(7)
The total flux from a point charge if 4πkq. Since the plain is infinite, half of the electric field lines
pass through this plane (not depending on d), that is half of the flux 2πkq.
1
Gauss’ law - cylindrical symmetry
Submitted by: I.D. 039023262
The problem:
Given an infinite cylinder charged ρ(~r) = ρ0
1. Find the electric field.
2. What should be the charge density ρ(~r) so that the electric filed inside the cylinder would be
zero?
The solution:
~ = Er r̂ . The volume element is dv = rdrdθdz
1. For cylindrical symmetry E
For r < R
Z
Z r
Er · 2πrh = 4πh
ρdv = 4πh
ρ2πr0 dr0 h
V
(1)
0
~ = 2πkρrr̂
E
(2)
For r > R
Z
Er · 2πrh = 4πh
Z
V
~ =
E
R
ρdv = 4πh
ρ2πr0 dr0 h
(3)
0
2πkρR2
r̂
r
(4)
2. We solve the problem in two ways - using integral and diferential versions of the Gauss’ law:
a. integral:
I
Z
~ = 4πk ρdv
~ · ds
E
(5)
Z r
E · 2πrh = 4πk
ρ(r0 ) · 2πr0 dr0 h
(6)
0
ρ(r) ∼
1
r
(7)
Checking:
Z r
ρ(r0 ) · 2πr0 dr0 h = 2πrh
(8)
0
~ = 4πkr̂ = const
E
(9)
b. differential:
~ = 4πkρ
div E
(10)
~ = const
~ =E
~ ⇒ div E
E
r
1
E
4πk
ρ =
⇒
=
r
r
r
~ = 4πkr̂ = const
E
(11)
(12)
(13)
1