#### Question

When 1⋅0 × 10^{12} 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.

#### Solution

As 1.0 × 10^{12 }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` .