Question

The figure shows two identical parallel plate capacitors connected to a battery with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. The ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric is $\frac{x}{5}$. Then find the value of x

Answer 1

Initially the charge on either capacitor i.e., $q_A$ or $q_B$ is CV coulomb. When dielectric is introduced, the new capacitance of either capacitor $C_1=KC=3C$

After the opening of switch S, the potential across capacitor A is V volts. Let the potential across capacitor B is $V_1$

$\therefore q_B=CV=C_1{V}_{1}orCV=3C{V}_1$

$\therefore V_1=\frac{V}{3}volts$

Initial energy of capacitor A = $\frac{1}{2}C{V}^2$

energy of capacitor B = $\frac{1}{2}C{V}^2$

$\therefore$ total energy $E_i=\frac{1}{2}C{V}^2+\frac{1}{2}C{V}^2=C{V}^2$

Final energy of capacitor A = $\frac{1}{2}\times \left(3C\right)V^2=\frac{3}{1}C{V}^2$

Final energy of capacitor B = $\frac{1}{2}\times \left(3C\right){\left(\frac{V}{3}\right)}^2=\frac{C{V}^2}{6}$

$\therefore \frac{E_i}{E_f}=\frac{C{V}^2}{\left(\frac{5}{3}\right)C{V}^2}=\frac{3}{5},x=3$