Ch24Word - Chapter 24 p 1 CHAPTER 24 Capacitance...

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Chapter 24 p. 1 CHAPTER 24 – Capacitance, Dielectrics, Electric Energy Storage 1. From Q = CV , we have 2500 µ C = C (950 V), which gives C = 2.6 F . 2. From Q = CV , we have 28.0 × 10 –8 C = (12,000 × 10 –12 F) V , which gives V = 23.3 V . 3. From Q = CV , we have 75 pC = C (12.0 V), which gives C = 6.3 pF . 4. The final potential on the capacitor will be the voltage of the battery. Positive charge will move from one plate to the other, so the charge that flows through the battery is Q = CV = (15.6 × 10 –6 F)(12 V) = 1.9 × 10 –4 C . 5. From Q = CV , we see that ? Q = C ? V ; 16 C = C (48 V – 28 V), which gives C = 0.80 F . 6. When the capacitors are connected, some charge will flow from C 1 to C 2 until the potential difference across the two capacitors is the same: V 1 = V 2 = V . Because charge is conserved, we have Q 0 = Q 1 + Q 2 . For the two capacitors we have Q 1 = C 1 V , and Q 2 = C 2 V . When we form the ratio, we get Q 2 / Q 1 = ( Q 0 Q 1 )/ Q 1 = C 2 / C 1 , which gives Q 1 = Q 0 C 1 /( C 1 + C 2 ). For Q 2 we have Q 2 = Q 0 Q 1 = Q 0 {1 – [ C 1 /( C 1 + C 2 )]}, thus Q 2 = Q 0 C 2 /( C 1 + C 2 ). We find the potential difference from Q 1 = C 1 V ; Q 0 C 1 /( C 1 + C 2 ) = C 1 V , which gives V = Q 0 /( C 1 + C 2 ). 7. We assume that the charge transferred is small compared to the initial charge on the plates so the potential difference between the plates is constant. The energy required to move the charge is W = qV . Thus the charge on each plate is Q = CV = C ( W / q ) = (16 × 10 –6 F)(25 J)/(0.20 × 10 –3 C) = 2.0 C . Because this is much greater than the charge moved, our assumption is justified.
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Chapter 24 p. 2 8. We find the initial charges on the capacitors: Q 1 = C 1 V 1 = (2.40 µ F)(880 V) = 2112 C; Q 2 = C 2 V 2 = (4.00 F)(560 V) = 2240 C. ( a ) When the capacitors are connected with positive plates together, some charge will flow from C 2 to C 1 until the potential difference across the two capacitors is the same: V 1 = V 2 = V . Because charge is conserved, we have Q = Q 1 + Q 2 = Q 1 + Q 2 = 2112 C + 2240 C = 4352 C. For the two capacitors we have Q 1 = C 1 V , and Q 2 = C 2 V . When we add these, we get Q 1 + Q 2 = Q = ( C 1 + C 2 ) V ; 4352 C = (2.40 F + 4.00 F) V , which gives V = 680 V . The charge on C 1 is Q 1 = C 1 V = (2.40 F)(680 V) = 1.63 × 10 3 C = 1.63 × 10 –3 C . The charge on C 2 is Q 2 = C 2 V = (4.00 F)(680 V) = 2.72 × 10 3 C = 2.72 × 10 –3 C . ( b ) When the capacitors are connected with opposite plates together and charge flows from C 2 to C 1 , the combination of positive and negative charges will result in the cancellation of some charge until the potential difference across the two capacitors is the same: V 1 = V 2 = V . Because charge is conserved, we have Q = Q 1 + Q 2 = Q 2 Q 1 = 2240 C – 2112 C = 128 C. For the two capacitors we have Q 1 = C 1 V , and Q 2 = C 2 V . When we add these, we get Q 1 + Q 2 = Q = ( C 1 + C 2 ) V ; 128 C = (2.40 F + 4.00 F) V , which gives V = 20 V . The charge on C 1 is Q 1 = C 1 V = (2.40 F)(20 V) = 48 C . The charge on C 2 is Q 2 = C 2 V = (4.00 F)(20 V) = 80 C . 9. For a parallel-plate capacitor, we find the area from C = Å 0 A / d ; 0.40 × 10 –6 F = (8.85 × 10 –12 C 2 /N · m 2 ) A /(4.0 × 10 –3 m), which gives A = 1.8 × 10 2 m 2 .
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Ch24Word - Chapter 24 p 1 CHAPTER 24 Capacitance...

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