Version One – Homework 1 – Juyang Huang – 24018 – Jan 16 ...

Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 20
For a conducting sphere, the charge is uniformly
distributed at the surface. Based on
Gauss’ law, the electric field on the surface of
a conducting sphere of radius R with charge
Q is
Q
E(r) = k e , where r ≥ R . (2)
r2 Thus on the surface r = R of the two
spheres,
E 2
E 1
=
k q 2
r 2 2
k q 1
(
r1
r 2 1
) 2
=
r 2
( 4 R
=
3 R
( 2 4
=
3)
= 16 9 .
) 2
044 (part 3 of 4) 10 points
Now “connect” the two spheres with a wire.
r 1
r 2
5. E 2
= 16 E 1 9
6. E 2
= 9
E 1 16
7. E 2
= 9 E 1 8
8. E 2
= 9
E 1 32
9. E 2
E 1
= 1
Explanation:
When the spheres are connected by a wire,
charge will flow from one to the other until
the potential on both spheres is the same.
As noted, V 2
V 1
= 1, defines equilibrium.
The spheres are connected by a wire and no
current is flowing (at equilibrium), therefore
the ends of the wire are at the same potential
V 2 = V 1 . (3)
For a conducting sphere, the charge is uniformly
distributed at the surface. Based on
Gauss’ law, on the surface of a conducting
sphere of radius R with charge Q is
Q
E(r) = k e , where r ≥ R , and
r2 V (r) = k Q r , where r ≤ R .
#1
q 1
q 2 #2
Thus on the surface r = R of the two
spheres,
There will be a flow of charge through the
wire until equilibrium is established.
What is the ratio of the electric fields E 2
E 1
at
the “surfaces” of the two spheres
1. E 2
E 1
= 4 3 correct
2. E 2
E 1
= 3 4
3. E 2
E 1
= 3 2
4. E 2
E 1
= 3 8
E 2
E 1
=
=
=
= r 1
r 2
k q 2
r2
2
k q 1
r1
2
k q 2 1
r 2 r 2
k q 1 1
r 1 r 1
V 2
1
r 2
V 1
1
r 1
, since V 1 = V 2
(4)