Homework 4 Solutions

Homework 4 Solutions
Problem 4.1: Square loop (based on Griffiths 2.4) Consider a square loop
with each side a length a carrying a uniform linear charge density λ.
a) Find the electric field at the center of the square.
Solution: The electric field to to each of the opposite sides will be in an
opposite direction, so the total electric field at the center of the square is
zero.
b) Find the work needed to bring a charge in from infinity to the center of
the square. (You may start with the electrostatic potential due to a single
finite line of charge.)
Solution: The work needed to bring a charge Q in from infinity to the
origin is just the charge times the electrostatic potential at the origin.
U = QV (0)
(1)
The electrostatic potential due to a finite line charge, in the plane that
bisects the line, is
!
√
a2 + 4r2 + a
(2)
V (r) = kλ ln √
a2 + 4r2 − a
!
√
2+1
V (l/2) = kλ ln √
(3)
2−1
Since there are four sides, we just end up with
U = 4kQλ ln
!
√
2+1
√
2−1
(4)
Problem 4.2: Three charges (Griffiths 2.32) Three charges are situated at
the corners of a square (side s). Two have charge −q and are located on opposite
corners. The third has charge +q and is opposite an empty corner.
a) How much work does it take to bring in another charge, +q, from far away
and place it at the fourth corner?
1
Solution: To find the work needed, I need to calculate the potential at
the final location of the fourth charge due to the presence of the other
charges. (The subscripts denote the charge that is creating the potential.)
W
=
q V (~r)
=
q (V1 + V2 + V3 )
k(−q) k(q) k(−q)
q
+√ +
s
s
2s
√ !
2
−k q 2
2−
s
2
=
=
The negative sign indicates that the charge is losing energy is being placed
at the corner of the square.
b) How much work does it take to assemble the whole configuration of four
charges?
Solution:
4
W =
1X
qi V (~ri )
2 i=1
The potential for charges on opposite corners is the same (here, the subscripts denote the total potential at the location of the charge due to the
presence of the other four charges):
VT 1 = VT 3
k q2
=
s
VT 2 = VT 4 =
W
=
=
−k q 2
s
√ !
2
2−
2
√ !
2
2−
2
1
[2(−q)V1 + 2(q)V2 ]
2
√ !
−(q) k q 2
(−q) k q 2
2
2−
+
s
2
s
k q2
= −2
s
√ !
2
2−
2
√ !
2
2−
2
The negative sign indicates that the system loses some energy in order to
be placed in the square.
2
Problem 4.3: Finite disk
a) Find the electrostatic potential everywhere along the axis of symmetry
due to a finite disk of charge with uniform (surface) charge density σ.
Solution: The disk can be viewed as consisting of lots of rings of radius
r0 and width dr0 . The charge on each such ring is
Q(r0 ) = σ(2πr0 dr0 )
and the potential for each such ring is, by part (a),
Q(r0 )
1
p
4π0 r0 2 + z 2
V (z) =
Adding up all the rings which make up a disk of radius R, we obtain
Z
V (z)
R
Z
2π
kσ
s0 dφ0 ds0
s0 2 + z 2
Z R
1
p
= 2πkσ
s0 ds0
2
0
0
s + z2
R
p
2
= kσ2π s0 + z 2 0
p
2
2
R +z −z
= 2πkσ
=
0
p
0
b) Find two nonzero terms in a series expansion of your answer to part (b)
for the value of the potential very far away from the disk.
Solution:
If z R, then
p
R2 + z 2
r
R2
z2
1 R2
1 R4
= z 1+
−
+
...
2 z2
8 z4
= z
1+
Inserting this into the result obtained in (b) leads to
σ p 2
2π R + z 2 − 2πz
V (z) =
4π0
2
σ
πR
πR4
=
−
+
...
4π0
z
4z 3
3
Problem 4.4: Potential of a cone 1 A conical surface (an empty ice-cream
cone) carries a uniform charge density σ. The height of the cone is a, as is the
radius of the top. Find the potential at the point in the center of the opening
of the cone), letting the potential at infinity be zero.
Solution: The cone that I’ll be considering has its vertex on the origin and
opens up on the positive z-axis. I’ll use cylindrical coordinates in my solution.
Starting with the general expression for calculating potential:
Z Z
k σ dA0
V (~r) =
|~r − ~r 0 |
then I’ll use what I know. For a cone with height h and base radius s, the
differential area element is:
s p
dA0 = 2 s2 + h2 z dz dφ
h
in this problem, the height and the radius of the base are the same value a,
so the differential area element is:
dA0 =
√
2 z 0 dz 0 dφ0
Using this differential area element, and rewriting the denominator is cylindrical coordinates, the potential becomes:
√
Z 2π Z a
2 k σ z 0 dz 0 dφ0
q
V (~r) =
0
0
r2 + r0 2 − 2rr0 cos(φ − φ0 ) + (z − z 0 )2
Now, simplifying the denominator, on the z-axis where the potential is evaluated, z = a, r = 0 and for our cone r0 = z 0 :
Z
V (z = a)
2π
Z
a
=
0
Z
0
2π
Z
=
0
0
a
√
2 k σ z 0 dz 0 dφ0
q
z 0 2 − +(a − z 0 )2
√
2 k σ z 0 dz 0 dφ0
p
2z 0 2 − 2zz 0 + a2
Using Maple or an integral table to evaluate the integral:
"
√
√ #
a 2
2+ 2
√
V (z = a) = k σ
ln
4
2− 2
1 based
on GEM 2.27
4