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

Therefore
Version One – Homework 1 – Juyang Huang – 24018 – Jan 16, 2008 32
C 12
′ =
2 κ
C 12 1 + κ .
4. U ′
063 (part 2 of 3) 10 points
of potential differences across
V 2
6. U ′
2 κ
1 + κ .
V 2 = Q 2
= V C 12
= V C 2 C 2 2 .
V 2 ′ = Q′ 2
C 2
′ = V C′ 12
C 2
′
= V 1+κ
κ C
κ C = V
1 + κ .
V 2
′
= 2
V 2 1 + κ .
064 (part 3 of 3) 10 points
of total energy stored in the
U
U = κ .
The ratio V ′
2
capacitor C 2 for the two cases is
1. V ′
2
V 2
=
2. V ′
2
V 2
= κ .
3. V ′
2
V 2
= 2
1 + κ . correct
4. V ′
2
V 2
= 1 + κ
2 κ .
5. V 2
′
= 1 + κ .
V 2 2
6. None of these.
Explanation:
For Case One,
For Case Two,
Therefore
The ratio U ′
capacitors for the two cases is
1. None of these
2. U ′
U = 2
1 + κ .
3. U ′
U = 1 + κ
2 κ .
5. U ′
U = 1 + κ
2
U =
Explanation:
For Case One,
For Case Two,
Therefore
keywords:
.
2 κ
1 + κ . correct
U = 1 2 C 12 V 2 .
U ′ = 1 2 C′ 12 V 2 .
U ′
U = C′ 12
C 12
=
2 κ
1 + κ .
Dielectric in a Capacitor 01
26:05, trigonometry, multiple choice, > 1 min,
wording-variable.
065 (part 1 of 1) 10 points
a) An isolated capacitor has a dielectric slab κ
between its plates.
b) The capacitor is charged by a battery.
c) After the capacitor is charged, the battery
is removed.
d) The dielectric slab is then moved half way
out of the capacitor.
e) Finally, the dielectric is released and is set
free to move on its own.
κ
The dielectric will
1. be pulled back into the capacitor. correct
2. remain in place.
κ