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Unformatted text preview: 1 QE , not QE
(c) dW # F dx # dU , so F # The reason for the difference is that E is the field due to both plates. If we consider the positive plate only and
calculate its electric field using Gauss’s law (Figure 24.29b):
!E + dA # !e0ncl
2 EA #
2!0 2!0 A
The force this field exerts on the other plate, that has charge !Q, is F #
24.30. IDENTIFY:! C # P0 A . The stored energy can be expressed either as Q2
2!0 A Q2
, whichever is more convenient
for the calculation.
SET UP:! Since d is halved, C doubles.
EXECUTE:! (a) If the separation distance is halved while the charge is kept fixed, then the capacitance increases and
the stored energy, which was 8.38 J, decreases since U # Q 2 2C. Therefore the new energy is 4.19 J.
(b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance
leads to a doubling of the stored energy to 16.8 J, using U # CV 2 2 , when V is held constant throughout. Capacitance and Dielectrics 24.31. 24-9 EVALUATE:! When the capacitor is disconnected, the stored energy decreases because of the positive work done by
the attractive force between the plates. When the capacitor remains connected to the battery, Q # CV tells us that the
charge on the plates increases. The increased stored energy comes from the battery when it puts more charge onto the
IDENTIFY and SET UP:! C # . U # 1 CV 2 .
EXECUTE:! (a) Q # CV # (5.0 $ F)(1.5 V) # 7.5 $ C . U # 1 CV 2 # 1 (5.0 $ F)(1.5 V)2 # 5.62 $ J
(b) U # 1 CV 2 # 1 C (Q / C ) 2 # Q 2 / 2C . Q # 2CU # 2(5.0 " 10!6 F)(1.0 J) # 3.2 " 10!3 C .
2 Q 3.2 " 10!3 C
# 640 V .
C 5.0 " 10!6 F
EVALUATE:! The stored energy is proportional to Q 2 and to V 2 .
V# 24.32. IDENTIFY:! The two capacitors are in series. 1
0 % C # . U # 1 CV 2 .
Ceq C1 C2 SET UP:! For capacitors in series the voltages add and the charges are the same.
(150 nF)(120 nF)
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This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.
- Fall '08