Chapter 24 HW Solutions

# Chapter 24 HW Solutions - PHYS 241 24.5.IDENTIFY C = Ch 24...

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PHYS 241 Ch. 24 Homework Solutions 24.5.I DENTIFY : ab Q C V = . 0 A C d = P . S ET U P : When the capacitor is connected to the battery, 12.0 V ab V = . E XECUTE : (a) 6 4 (10.0 10 F)(12.0 V) 1.20 10 C 120 C ab Q CV μ - - = = × = × = (b) When d is doubled C is halved, so Q is halved. 60 C Q = . (c) If r is doubled, A increases by a factor of 4. C increases by a factor of 4 and Q increases by a factor of 4. 480 C. Q = E VALUATE : When the plates are moved apart, less charge on the plates is required to produce the same potential difference. With the separation of the plates constant, the electric field must remain constant to produce the same potential difference. The electric field depends on the surface charge density, σ . To produce the same , more charge is required when the area increases. 24.11. I DENTIFY and S ET U P : Use the expression for C / L derived in Example 24.4. Then use Eq.(24.1) to calculate Q. E XECUTE : (a) From Example 24.4, ( 29 0 2 ln / b a C L r r π = P ( 29 ( 29 12 2 2 11 2 8.854 10 C / N m 6.57 10 F/m 66 pF/m ln 3.5 mm/1.5 mm C L - - × = = × = (b) ( 29 ( 29 11 10 6.57 10 F/m 2.8 m 1.84 10 F. C - - = × = × ( 29 ( 29 10 3 11 1.84 10 F 350 10 V 6.4 10 C 64 pC Q CV - - - = = × × = × = The conductor at higher potential has the positive charge, so there is +64 pC on the inner conductor and 64 - pC on the outer conductor. E VALUATE : C depends only on the dimensions of the capacitor. Q and V are proportional. 24.15. I DENTIFY : Replace series and parallel combinations of capacitors by their equivalents. In each equivalent network apply the rules for Q and V for capacitors in series and parallel; start with the simplest network and work back to the original circuit. S ET U P : Do parts (a) and (b) together. The capacitor network is drawn in Figure 24.15a. 1 2 3 4 400 F C C C C = = = = 28.0 V ab V = Figure 24.15a E XECUTE : Simplify the circuit by replacing the capacitor combinations by their equivalents: 1 2 and C C are in series and are equivalent to 12 C (Figure 24.15b). 12

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Chapter 24 HW Solutions - PHYS 241 24.5.IDENTIFY C = Ch 24...

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