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.
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.
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` .