The figures below show 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 capacitor is filled with dielectric k = 2. The ratio of the Electrostatic Energy stored in both the capacitors before and after the introduction of the slab is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.8
With S open, the capacitor on the right is isolated, so its charge stays constant. The capacitor across the battery is still connected to it, so its potential difference stays constant. Energy is \( \frac{1}{2}CV^{2} \) for the connected one and \( \frac{Q^{2}}{2C} \) for the isolated one. Filling with a dielectric of \( k = 2 \) doubles the capacitance of each (shown in the figure as the shaded plates).
Before (S closed, both at the battery voltage V)
\( \displaystyle U_i = \frac{1}{2}CV^{2} + \frac{1}{2}CV^{2} = CV^{2} \)
After
\( \displaystyle U_1 = \frac{1}{2}(2C)V^{2} = CV^{2} \qquad (\text{battery connected, } V \text{ constant}) \)
\( \displaystyle U_2 = \frac{Q^{2}}{2(2C)} = \frac{(CV)^{2}}{4C} = \frac{CV^{2}}{4} \qquad (\text{isolated, } Q = CV \text{ constant}) \)
\( \displaystyle U_f = CV^{2} + \frac{CV^{2}}{4} = \frac{5}{4}CV^{2} \)
\( \displaystyle \frac{U_i}{U_f} = \frac{CV^{2}}{\frac{5}{4}CV^{2}} = \frac{4}{5} = 0.8 \)