This preview shows pages 1–3. Sign up to view the full content.
Physics 217  Homework 3
Solutions
Ruxandra Bondarescu
February 21, 2005
Problem 1. Purcell 2.5
The potential for a basketball size sphere is
ϕ
= 1000 volts = 3
.
3 statvolts. This potential is the
same as the potential of a point charge
ϕ
=
Q/R
, where
Q
is the radius of the sphere and R is the
radius of the basketball. The charge per
cm
2
is
Q
4
πR
2
=
ϕ
4
πR
=
3
.
3 statvolts
4
π
20 cm
(1)
where I assumed that the radius of a basketball is
R
= 20
cm
. The number of electrons is then
given by Eq. (1) divided by the charge of an electron
#electrons
cm
2
=
3
.
3 statvolts
4
π
(20 cm) (5 10

10
esu)
= 2
.
62 10
7
(2)
So, the number of extra electrons per
cm
2
is about 3
×
10
7
.
Problem 2. Purcell 2.13
∇ ×
E
= ˆ
z
(
∂E
y
∂x

∂E
x
∂y
) = ˆ
z
(6
x

6
x
) = 0
(3)
which checks that the Feld given in problem 2.1 is a proper electrostatic Feld.
∇ ·
E
=
∂E
x
∂x
+
∂E
y
∂y
= 6
y
+ (

6
y
) = 0
(4)
This shows that there are no charges enclosed.
Problem 3. Purcell 2.18
Method 1.
The potential di±erence Δ
ϕ
=

i
center
edge
E
·
d
l
. We discretize the cylinder in rings of
charge. The Feld due to an arbitrary ring of charge
dq
of width
dx
is a ring of charge on axis is
dE
=
dE
x
ˆx
=
xdq
(
x
2
+
a
2
)
3
/
2
ˆx
(5)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document The total electric feld at a point
x
0
on axis is the sum over all such rings
E
(
x
0
) =

i
x
0
0
2
πaσxdx
(
x
2
+
a
2
)
3
/
2
+
i
b

x
0
0
2
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 08/07/2008 for the course PHYS 2217 taught by Professor Leclair, a during the Spring '06 term at Cornell University (Engineering School).
 Spring '06
 LECLAIR, A
 Charge, Work

Click to edit the document details