A parallel-plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (V) as ε=α V where α=3 V -1. A similar capacitor with no dielectric is charged to V 0=52 V. It is then connected to the uncharged capacitor with the dielectric. What is the final voltage (in V) on the capacitor?

Question

A parallel-plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (V) as ε=α V where α=3 V -1. A similar capacitor with no dielectric is charged to V 0=52 V. It is then connected to the uncharged capacitor with the dielectric. What is the final voltage (in V) on the capacitor? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

On connecting the two given capacitors, let the final voltage be

V

.
If the capacitance of capacitor without the dielectric is

C

then,

q_{1} = C V

The other capacitor with dielectric has capacitance

εC

.

∴ q_{2} = εC V

= α C V^{2} \dots . [∵ ε = α V]

The initial charge on the capacitor (without dielectric) that was charged,

q_{0} = C V_{0}

from conservation of charge,

q_{0} = q_{1} + q_{2}

∴ C V_{0} = C V + α C V^{2}

∴ α V^{2} + V - V_{0} = 0

∴ V = \frac{- 1 \pm 1 + 4 α V _{0}}{2 α}

= \frac{- 1 \pm 1 + 4 \times ( 3 ) \times ( 52 )}{2 \times ( 3 )}

= \frac{- 1 \pm 625}{2 \times ( 3 )}

As

V

is positive,

∴ V = \frac{625 - 1}{6}

= \frac{24}{6} = 4 V