Question

An air filled parallel plate capacitor charged to potential $V_1$ is connected to an uncharged identical parallel plate capacitor with dielectric constant $K$. The common potential is $V_2$. The value of $K$ is

• A. $\frac{V_1-V_2}{V_1+V_2}$
• B. $\frac{V_1}{V_1-V_2}$
• C. $\frac{V_1-V_2}{V_2}$
• D. $\frac{V_1-V_2}{V_1}$

Answer 1

The correct option is C $\frac{V_1-V_2}{V_2}$

Using the formula

$V_2=\frac{CV+C_1^{\prime }}{C+C^{\prime }}⟶\left(i\right)$

Charges of 1st Capacitor $Q_1=0$

$CV=0⟶\left(ii\right)$

$C_1=\frac{K{ϵ}_{o}A}{d}⟶\left(iii\right)$

$C^{\prime }=\frac{{ϵ}_{o}A}{d}⟶\left(iv\right)$

By Equation $\left(i\right),\left(ii\right),\left(iii\right),\left(iv\right)$

$V_2=\frac{0+\frac{{ϵ}_{o}A}{d}\left(V_1\right)}{\frac{K{ϵ}_{o}A}{d}+\frac{{ϵ}_{o}A}{d}}$

$V_2=\frac{V_1}{K+1}$

$\Rightarrow K=\frac{V_1}{V_2}-1=\frac{V_1-V_2}{V_2}$