Problem%20Set%203%20Solutions

Problem%20Set%203%20Solutions - 16. We note that the...

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16. We note that the voltage across C 3 is V 3 = (12 V – 2 V – 5 V ) = 5 V. Thus, its charge is q 3 = C 3 V 3 = 4 µ C. (a) Therefore, since C 1 , C 2 and C 3 are in series (so they have the same charge), then C 1 = 4 µ C 2 V = 2.0 µ F . (b) Similarly, C 2 = 4/5 = 0.80 µ F.
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24. For maximum capacitance the two groups of plates must face each other with maximum area. In this case the whole capacitor consists of ( n – 1) identical single capacitors connected in parallel. Each capacitor has surface area A and plate separation d so its capacitance is given by C 0 = ε 0 A / d . Thus, the total capacitance of the combination is () 12 2 2 4 2 0 12 0 3 1 (8 1)(8.85 10 C /N m )(1.25 10 m ) 12 . 2 8 1 0 F . 3.40 10 m nA Cn C d −− −× × =− = = = × ×
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26. The charges on capacitors 2 and 3 are the same, so these capacitors may be replaced by an equivalent capacitance determined from 11 1 23 CC C eq =+= + . Thus, C eq = C 2 C 3 /( C 2 + C 3 ). The charge on the equivalent capacitor is the same as the charge on either of the two capacitors in the combination and the potential difference across the equivalent capacitor is given by q 2 / C eq . The potential difference across capacitor 1 is q 1 / C 1 , where q 1 is the charge on this capacitor. The potential difference across the combination of capacitors 2 and 3 must be the same as the potential difference across capacitor 1, so q 1 / C 1 = q 2 / C eq . Now some of the charge originally on capacitor 1
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This note was uploaded on 05/04/2010 for the course PHYS 11b taught by Professor Mobius during the Spring '09 term at Harvard.

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Problem%20Set%203%20Solutions - 16. We note that the...

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