Question

A system of two conductors is placed in air and they have net charges of +80µC and −80µC which causes a potential difference of 16V between them.
1. Find the capacitance of the system.
2. If the air between the capacitor is replaced by a dielectric medium of di electric constant K, 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:

The capacitance of a system is determined by its geometry and the dielectric constant of the medium between the conductors, and it is independent of the amount of charge.

Answer 1

% Option (1) The capacitance \( C \) of the system can be found using the formula for the capacitance between two conductors: \[ C = \frac{Q}{V} \] where \( Q = 80 \, \mu C = 80 \times 10^{-6} \, C \) and \( V = 16 \, \text{V} \). Therefore: \[ C = \frac{80 \times 10^{-6}}{16} = 5 \times 10^{-6} \, \text{F} = 5 \, \mu F \] % Option (2) If the air is replaced by a dielectric medium of dielectric constant \( K \), the capacitance increases by a factor of \( K \). The new capacitance \( C' \) will be: \[ C' = K \cdot C \] Thus, the new potential difference will be reduced by a factor of \( K \) as the capacitance increases. The new potential difference \( V' \) is: \[ V' = \frac{V}{K} \] % Option (3) If the charges on the conductors are doubled to \( +160 \, \mu C \) and \( -160 \, \mu C \), the capacitance will not change because the capacitance of a system depends only on the geometry and the dielectric medium between the conductors, not on the charge. Therefore, the capacitance remains \( 5 \, \mu F \). However, the potential difference will double due to the increased charge.