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 capacitance \( C \) of the system can be calculated using the formula for capacitance: \[ C = \frac{Q}{V} \] where:
\( Q = 80 \, \mu C = 80 \times 10^{-6} \, C \),
\( V = 16 \, \text{V} \).
Therefore, the capacitance is: \[ C = \frac{80 \times 10^{-6}}{16} = 5 \, \mu F \]

(ii) When the dielectric constant \( k = 3 \) is inserted, the capacitance increases by a factor of \( k \). The new capacitance \( C' \) becomes: \[ C' = kC = 3 \times 5 \, \mu F = 15 \, \mu F \] Since \( Q = C'V' \), and the charge \( Q \) remains the same, the new potential difference \( V' \) is: \[ V' = \frac{Q}{C'} = \frac{80 \times 10^{-6}}{15 \times 10^{-6}} = 5.33 \, \text{V} \]

(iii) The capacitance of the system depends only on the geometry of the conductors and the dielectric constant of the medium between them, not the charges. Therefore, if the charges are doubled, the capacitance remains unchanged. The capacitance will still be \( 5 \, \mu F \), because capacitance is independent of the charge.