Download Electric force

Electric force
Submitted by: I.D. 061110185
The problem:
Given 4 charges arranged at the corners of a square with sides of a length 2a (as on the picture)
find the force acting on the charge Q at some point on the x-axis.
2
sign(x)
ŷ.
Show that for |x| a F~ = 3Qqa
πε0 ·
x4
The solution:
Coulomb law
F~
= kq1 q2
k =
~r
|r|3
(1)
c2
1
ε0 = 8.85 · 10−12
4πε0
N m2
(2)
We use the superposition principle and find the forces acting from each one of the charges. We
define ~ri as a vector pointing from the charge qi to the charge Q. Then
~r1 = (x − a)~i − aĵ
~r2 = (x + a)~i − aĵ
(3)
~r3 = (x + a)~i + aĵ
~r4 = (x − a)~i + aĵ
(5)
(4)
(6)
where the the charges qi are counted from the top right one in the counterclockwise direction.
Therefore, the force is
F~
= F~(x,0,0) = F~1 + F~2 + F~3 + F~4 =
(7)
!
(x + a)~i − aĵ
(x + a)~i + aĵ
(x − a)~i + aĵ
(x − a)~i − aĵ
+
−
+
3/2
3/2
3/2
2
2
2
2
2
2
((x − a) + a )
((x + a) + a )
((x + a) + a )
((x − a)2 + a2 )3/2
1
1
ĵ
(8)
= 2akqQ −
+
((x + a)2 + a2 )3/2 ((x − a)2 + a2 )3/2
= kqQ −
For |x| a
1
((x ± a)2 + a2 )3/2
=
'
1
1
=
2a
3
(x2 ± 2ax + 2a2 )3/2
|x| (1 ± x +
3a
1
1
'
1∓
3/2
|x|3
x
|x|3 (1 ± 2a
x)
1
2a2 3/2
)
x2
(9)
(10)
Then
3Qqa2 sign(x)
·
ŷ
F~ =
πε0
x4
(11)
which is the the force acting by a quadrupole.
2
Electric force
The problem:
An infinite wire charged uniformly with the charge density λ is bent as shown on the picture. What
is the force acting on a charge q which is put into the center of the half-circle?
The solution:
We divide the problems to 2 parts: (1) finding the force acting from the straight line and (2) finding
the force acting from the half circle.
The infinidecimal charge is dq = λdl.
For the straight line:
~r0 = xî + Rĵ
kQdq
kQλxdx
= −
3 x = −
3
(x2 + R2 ) 2
(x2 + R2 ) 2
kλRdx
= −
3
(x2 + R2 ) 2
Z∞
kQλxdx
2λQk
= −2
3 = −
R
(x2 + R2 ) 2
~r = 0 ,
dFx
dFy
Fx
(1)
(2)
(3)
(4)
0
where the last integral could be solved (for example) by the substitution x = cos α. The ycomponents of the force are canceled since there are 2 wires, and the factor of 2 in the last equation
also comes from this fact.
For the half circle:
~r0 = R cos θî + R sin θĵ
kQdq
kQλR2 dθ
dFx =
R
cos
θ
=
cos θ
R3
R3
2λQk
Fx =
R
(5)
(6)
(7)
Therefore,
X
F~ = 0
(8)
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