Question

A system of two conductors is placed in air and they have net charge of \( +80 \, \mu C \) and \( -80 \, \mu C \) which causes a potential difference of 16 V between them.
(1) Find the capacitance of the system.
(2) If the air between the capacitor is replaced by a dielectric medium of dielectric constant 3, what will be the potential difference between the two conductors?
(3) If the charges on two conductors are changed to +160µC and −160µC, will the capacitance of the system change? Give reason for your answer.

Hint:

Capacitance depends on the geometry of the conductors and the dielectric material, not the amount of charge.

Answer 1

The system's capacitance \( C \) is determined by the formula \( C = \frac{Q}{V} \), with \( Q = 80 \, \mu C = 80 \times 10^{-6} \, C \) and \( V = 16 \, \text{V} \). This yields a capacitance of \( C = \frac{80 \times 10^{-6}}{16} = 5 \, \mu F \).

(ii) Introducing a dielectric with constant \( k = 3 \) increases the capacitance by a factor of \( k \), resulting in a new capacitance \( C' = kC = 3 \times 5 \, \mu F = 15 \, \mu F \). Since \( Q = C'V' \) and \( Q \) is constant, the new potential difference is \( V' = \frac{Q}{C'} = \frac{80 \times 10^{-6}}{15 \times 10^{-6}} = 5.33 \, \text{V} \).

(iii) The capacitance of the system is solely dependent on the conductor geometry and the dielectric constant of the intervening medium, not the charges. Therefore, doubling the charges does not alter the capacitance, which remains \( 5 \, \mu F \) as capacitance is independent of charge.