Download Electric Potential and Energy

Electric Potential and Energy
Submitted by: I.D. 039033345
The problem:
How much energy is needed to create the following configuration?
The solution:
Let φi be the potential at the position of the charge qi as a result of all the other charges.
φi =
X kqj
i6=j
(1)
rj
Numbering the charges from the left upper corner in the clockwise direction we have
kq kq
kq
q1 φ1 = q − −
+√
a
a
2a
kq
kq kq
q2 φ2 = −q
+
−√
a
a
2a
q3 φ1 = q1 φ1
(2)
(3)
(4)
q4 φ4 = q2 φ2
kq 2 √
1X
U =
qi φi =
( 2 − 4) ⇒ U < 0
2
a
(5)
U <0
(7)
(6)
i
Then
From here we can conclude that the work needed to create this system is negative, therefore we
need to invest energy in order to dissolve the system.
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
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