Two conductors, insulated from each other, are charged by transferring electrons from one conductor to the other. After 2.5 × 1012 electrons have been transferred, the potential difference between the conductors is 12 V. What is the capacitance of the system?

33 nF

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27 nF

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87 nF

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108 nF

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Solution

The correct option is A 33 nF
$Q = 2.5 \times 10^{12} \times 1.6 \times 10^{- 19} Q = 4.0 \times 10^{- 7} C C = \frac{Q}{V} = \frac{4.0 \times 10^{- 7}}{12} = 33 n F C a p a c i t a n c e = 33 n F$

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Two conductors, insulated from each other, are charged by transferring electrons from one conductor to the other. After 2.5 × 1012 electrons have been transferred, the potential difference between the conductors is 12 V. What is the capacitance of the system?

The correct option is A 33 nFQ=2.5×1012×1.6×10−19Q=4.0×10−7CC=QV=4.0×10−712=33nFCapacitance=33nF

The correct option is A 33 nF
$Q = 2.5 \times 10^{12} \times 1.6 \times 10^{- 19} Q = 4.0 \times 10^{- 7} C C = \frac{Q}{V} = \frac{4.0 \times 10^{- 7}}{12} = 33 n F C a p a c i t a n c e = 33 n F$