When 1.0 × 1012 electrons are transferred from one conductor to another, a potential difference of 10V appears between the conductors. Calculate the capacitance of the two-conductor system.
Given:
n=1.0 × 1012
V= 10 V
Formula used:
1. Charge can be given by the formula
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Where Q→ Charge on the capacitor
n → number of the electrons
e → electric charge of an electron =![]()
2. The capacitance of a parallel plate Capacitor is given by
![]()
Where Q → charge on the capacitor
V → Voltage or potential difference
Putting the values in the formula 1, we get
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Substitute Q and C in Formula 2), we get
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Thus, the capacitance of the parallel plate capacitor is ![]()
So, if 1.0 × 1012 electrons are transferred between two conductors the capacitance of the parallel plate capacitor is
F when a potential difference is 10V