The figure shows two identical parallel plate capacitors A and B of capacitances C connected to a battery. The key K is initially closed. The switch is now opened and the free spaces between the plates of the capacitors are filled with a dielectric constant 3. Then which of the following statement(s) is/are true ?
A capacitor that stays connected to a battery keeps its potential difference fixed, while an isolated capacitor keeps its charge fixed. Filling the gap with a dielectric of constant 3 multiplies the capacitance by 3. In the circuit, A is permanently across the battery and the switch K joins B in parallel with it.
Step 1 — With K closed both capacitors have voltage V, so \(U=2\times\dfrac{1}{2}CV^{2}=CV^{2}\).
Step 2 — When K is opened, B is cut off carrying charge CV, and A remains on the battery at voltage V. Both are then filled with the dielectric, so each capacitance becomes 3C.
Step 3 — For A (voltage fixed): \(U_{A}=\dfrac{1}{2}(3C)V^{2}=\dfrac{3}{2}CV^{2}\).
Step 4 — For B (charge fixed): \(U_{B}=\dfrac{(CV)^{2}}{2(3C)}=\dfrac{1}{6}CV^{2}\).
Step 5 — Total energy after opening: \(U=\dfrac{3}{2}CV^{2}+\dfrac{1}{6}CV^{2}=\dfrac{5}{3}CV^{2}\).
The statement that B stores no charge is wrong, since B keeps the charge CV it had when K was opened. The value 3/2 CV2 is the energy of A, not of B, which holds only CV2/6. The total of CV2 with K closed and 5/3 CV2 after opening both hold.
When the switch is closed, total energy stored in the two capacitors is CV2. and When the switch is opened, total energy stored in two capacitors is \(\dfrac{5}{3}\) CV2. options are correct.
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