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SECTION 26.4 • Energy Stored in a Charged Capacitor 809 the problem, note that the charges on the left-hand plates The negative sign for Q 2i is necessary because the charge on the left plate of capacitor C 2 is negative. The total charge Q in the system is After the switches are closed, the total charge Q in the 1f and Q 2f . Because the system is isolated, The charges redistribute until the potential difference is the f . To satisfy this requirement, the charges on the capacitors after the switches are closed are Dividing the first equation by the second, we have Combining Equations (2) and (3), we obtain Using Equations (3) and (4) to find Q 1f in terms of Q, we have Finally, using Equation 26.1 to find the voltage across each As noted earlier, !V 1f # ! V 2f # ! V f . To express !V f in terms of the given quantities C 1 , C 2 , and !V i , we substitute the value of Q from Equation (1) into either Equation (6) or (7) to obtain
$ C 1 $ C
2 C 1 & C
2 %
∆V i ∆V f # (7)
∆V 2f # Q
2f C
2 # Q [C
2 /(C 1 & C
2 )] C
2 # Q C 1 & C
2 (6)
∆V 1f # Q
1f C 1 # Q [C 1 /(C 1 & C
2 )] C 1 # Q C 1 & C
2
# Q
$ C
1 C 1 & C
2 % (5)
Q
1f
# C 1 C
2 Q
2f # C 1 C
2 Q
$ C
2 C 1 & C
2 % (4)
Q
2f # Q
$ C
2 C 1 & C
2 % Q # Q
1f & Q
2f # C 1
C 2 Q
2f & Q
2f # Q
2f $ 1 & C
1 C
2 % (3)
Q
1f # C 1 C
2
Q
2f Q
1f # C 1
∆V f
and
Q
2f # C
2 ∆V f (2)
Q # Q
1f & Q
2f (1)
Q # Q
1i & Q
2i # (C 1 $ C
2 ) ∆V i Q
1i # C 1
∆V i
and
Q
2i # $ C
2
∆V i (B) Find the total energy stored in the capacitors before and after the switches are closed and the ratio of the final energy Solution Before the switches are closed, the total energy After the switches are closed, the total energy stored in the Using the results of part (A), we can express this as Therefore, the ratio of the final energy stored to the initial To finalize this problem, note that this ratio is less than unity, What If? What if the two capacitors have the same capaci- tance? What would we expect to happen when the switches Answer The equal-magnitude charges on the two capaci- Let us test our results to see if this is the case mathemati- cally. In Equation (1), because the charges are of equal mag- 1f # Q 2f # 0, consis- tent with our prediction. Furthermore, Equations (6) and 1f # ! V 2f # 0, which is consistent with uncharged capacitors. Finally, if C 1 # C 2 , Equation (8) shows us that U f # 0, which is also consistent with uncharged capacitors. $ C
1 $ C
2 C
1 & C
2 % 2 # (8)
U f U i # 1 2
(C
1 $ C
2 ) 2 ( ∆V i ) 2 /(C 1 & C
2 ) 1 2 (C 1 & C 2 )( ∆V i ) 2
1 2
(C
1 $ C
2 ) 2 ( ∆V i ) 2 (C
1 & C
2 ) U f # U f # 1 2 C
1 ( ∆V f
) 2 & 1 2 C
2 ( ∆V f
) 2 # 1 2 (C
1 & C
2 )(!V f
) 2
1 2 (C 1 & C
2 )( ∆V i ) 2 U i #
1 2 C
1 ( ∆V i ) 2 & 1 2 C
2 ( ∆V i ) 2 # At the Interactive Worked Example link at http://www.pse6.com, explore this situation for various initial values of the volt- |