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 2
The correct answer is Option (2) → 0.75
Initial Condition (Switch closed):
$C_{\text{total initial}}=C+C=2C$
Energy stored:
$U_{\text{initial}}=\frac{1}{2}.C_{\text{total}}.V^2=\frac{1}{2}.2C.V^2=CV^2$
Final Condition:
Switch is opened → charge is now fixed on each capacitor.
One capacitor is filled with dielectric of $k=2$, the other remains unchanged.
Charge remains same on each capacitor (as switch is open, no current can flow).
Let's calculate new energy in both:
Capacitor 1 (dielectric inserted):
- Capacitance increases to $2C$
- Charge remains $Q=CV$
- Energy = $U_1=\frac{Q^2}{2C}=\frac{(CV)^2}{2.2C}=\frac{CV^2}{4}$
Capacitor 2 (unchanged):
- Capacitance = $C$
- Charge = $CV$
- Energy = $U_2=\frac{(CV)^2}{2C}=\frac{CV^2}{2}$
Total final energy:
$U_{\text{final}}=\frac{CV^2}{4}+\frac{CV^2}{2}=\frac{3CV^2}{4}$
Energy Ratio:
$\frac{U_{\text{final}}}{U_{\text{initial}}}=\frac{\frac{3CV^2}{4}}{CV^2}=\frac{3}{4}$
= 0.75