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Question
When 1⋅0 × 1012 electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Calculate the capacitance of the two-conductor system.
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Solution
As 1.0 × 1012 electrons are transferred from one conductor to another, the conductor to which the electrons are transferred becomes negatively charged and the other conductor becomes positively charged.
Now ,
Magnitude of the net charge on each conductor, `Q = (1.0 xx 10^12) xx (1.6 xx 10^-19) C = 1.6 xx 10^-7 C`
Magnitude of the potential difference between the conductors, V = 10 V
The capacitance C is the ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between the conductors.
`C = Q/V`
⇒ `C = (1.6 xx 10^-7)/10 = 1.6 xx 10^-8 F`
Hence, the value of the capacitance of the given two conductor systems is `1.6 xx 10^-8 F` .
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