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 $V$.
$\therefore U_i=\frac{1}{2}C{V}^2+\frac{1}{2}C{V}^2$
$=\frac{1}{2}(2C)V^2=C{V}^2$
When the switch $S$ is opened and dielectric is introduced, the capacitance of each becomes $3C$. The potential of $A$ remains constant at $V$ and the charge on $B$ remains fixed.
$\therefore U_f=\frac{1}{2}(3C)V^2+\frac{1}{2}\frac{(VC{)}^2}{3C}$
$=\frac{3}{2}C{V}^2+\frac{1}{6}C{V}^2=\frac{5}{3}C{V}^2$
$\therefore \frac{U_i}{U_f}=\frac{C{V}^2}{\frac{5}{3}C{V}^2}=\frac{3}{5}$