Question

A parallel plate capacitor with air as dielectric is charged to a potential 'V' using a battery. Removing the battery, the charged capacitor is then connected across an identical uncharged parallel plate capacitor filled with wax of dielectric constant 'K' the common potential of both the capacitor is

• A. V volts
• B. kV volts
• C. (k + 1) V volts
• D. $\frac{V}{k+1}\text{volts}$

Answer 1

Answer: (D)

Capacitance with air as dielectric $C={\epsilon }_0\frac{A}{d}$

For identical Capacitor Capacitance with wax as dielectric $C^{\prime }=\epsilon \frac{A}{d}$

$\therefore C^{\prime }=KC,whereK=\frac{\epsilon }{{\epsilon }_0}$

When $Cand{C}^{\prime }$ are connected in parallel charge is distributed and the potential difference is equal for the two capacitors

Resultant capacitance $C_R=C+C^{\prime }$ and the total charge $Q=CV$

$\therefore$ Common \Potential \V=\dfrac{Q}{C_R}=\dfrac{CV}{C+C'}$

$\therefore V=\frac{CV}{C+KC}=\frac{V}{K+1}$