Question

The figure shows two identical parallel plate capacitors connected to a battery with the 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) $3$. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

• A. $\frac{2}{3}$
• B. $\frac{5}{3}$
• C. $\frac{3}{5}$
• D. $\frac{3}{2}$

Answer 1

Answer: (C)

When the switch $S$ is closed, the two capacitors in parallel will be charged by the same potential difference $V$.
$\therefore U_{i}=\frac{1}{2} C V^{2}+\frac{1}{2} C V^{2}$
$=\frac{1}{2}(2 C) V^{2}=C V^{2}$
When the switch $S$ is opened and dielectric is introduced, the capacitance of each becomes $3 C$. The potential of $A$ remains constant at $V$ and the charge on $B$ remains fixed.
$\therefore U_{f}=\frac{1}{2}(3 C) V^{2}+\frac{1}{2} \frac{(V C)^{2}}{3 C}$
$=\frac{3}{2} C V^{2}+\frac{1}{6} C V^{2}=\frac{5}{3} C V^{2}$
$\therefore \frac{U_{i}}{U_{f}}=\frac{C V^{2}}{\frac{5}{3} C V^{2}}=\frac{3}{5}$