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PHY 102 Electricity and Magnetism
Spring ’11
Dr. Pervez Hoodbhoy
Solutions to Assignment 3 (Continued)
1. Capacitance of a sphere is given by C = 4π0 R
Hence, initially the two potentials are: V1 =
Q
C1
=
Q
4π0 R1
and V2 =
Q
C2
=
Q
4π0 R2
Now, as R2 > R1 , V2 > V1 , and hence charge ∆Q flows from the R2 sphere to the R1
sphere till the potentials are the same.
Hence, in equilibrium, the charges are: Q1 = Q + ∆Q and Q2 = Q − ∆Q.
Equating the two final potentials, we get:
Q+∆Q
Q−∆Q
= 4π
4π0 R1
0 R2
1
1
⇒ ∆Q(1 + R
) = (R
− 1)Q
R2
R2
Hence, the charge flown, ∆Q =
−R1
+1)
R2
R1
(1+ R )
2
Q(
2. Note that the annul axis mentioned in this question is perpendicular to the plane of
the disc, that is, we want to find the potential a perpendicular distance y from the center
of the disc. [The case is similar to Chp 25, Example 10.]
So for part (a), let’s first compute the charge density:
Q
4Q
σ = π(R2 −R
2 /4) = 3πR2
1 dq
The potential dV = 4π
= k dqr
0 r
p
Here, r = x2 + y 2 where x goes from R/2 to R.
Taking a small segment of the disc, dA = 2πxdx, charge dq = σ2πxdx
Hence, dV = 2πkσ √ 2a 2 dx
x +y
RR
Therefore, potential V = 2πkσ R/2 √ x2 2 dx
x +y
This is the integral that we have to compute, using, of course, some (trigonometric) substitution. So let us define an angle θ, such that tan θ = xy
θ
So x = y cos
so dx = − (sinyθ)2 dθ
sin θ
Rθ
V = 2πkσ θ12 q yx2 dθ
x
+1
x2
R θ2
−y
V = 2πkσ θ1 (sin θ)2 sec θ dθ
= 2πkσ(−y)[csc θ2 − csc θ1 ] After some simple calculation, this gives, by plugging in
1
values of cosecant
of θ at the
p two limits, and using the expressions for k and σ:
p
2Q
2
2
V = 3π0 R2 [ y + R /4 − y 2 + R2 ]
For part (b), the Electric field can be simply found by taking the gradient of the potential: E = − dV
dy
3. We first want to find the Electric Field (or the potential, either method will work)
due to the first rod, at a point P, which is a fixed distance x from the rod, as shown in
the question:
Q
l
The charge density is given by λ =
Defining our length variable y such that y goes from 0 to l, and the distance from P
goes from x to x + l, we get, for a small segment dy:
dq = λdy
dy
= kλ (x+y)
dE = k dq
2
r2
R l dy
1
Hence, kλ 0 (x+y)2 = −kλ( x+l
= x1 )
Now, we have the Electric field due to the first rod as a function of the distance x from
the edge of the rod. This determines the work done, and hence the potential, in bringing
the second rod to the given position.
Hence,
Z
Z x+l
1
1
(
= )dx
V = − E.ds = kλ
x+l
x
x
kQ
x + 2l
x+l
V =
[ln(
) − ln
]
4π0 l
x+l
x
4. Every charge - charge interaction results in a contribution to the electric energy. This
means 30 interactions. Consider the positive charge in the top left√corner. It interacts
with 3 negative
charges at distance d, 3 positive charges at distance 2d and one negative
√
2
charge at d + 2d2 . Hence its electric energy is:
UA = −3
kq 2
kq 2
kq 2
+ 3√ − √
d
d2 + 2d2
2d
Then the total energy is just 6UA .
5. Since the particles are identical we should make use of symmetry.
U =3
k(2e)(2e)
3.0 × 10− 15
6. The area is split equally into two so A =
A
.
2
So simply C =
A = πr1 L + πr2 L
d = r2 − r1
Solving these two simultaneously gives us
r1 =
A − πdL
2πL
2
o A
2d
b)
r2 =
A + AdL
2πL
C=
2πL
ln(r2 /r1 )
Then the capacitance is
3