Question

The figure shows two similar capacitors connected to a battery. Initially, switch was closed and when opened, the capacitors were filled with a dielectric of dielectric constant $3$. The potential difference across capacitor B is:

• A. $6V$
• B. $3V$
• C. $\frac{3}{2}V$
• D. $2V$

Answer 1

The correct option is D $2V$

When switch is closed, both the capacitors are in parallel.
Charge on both capacitors: $q_A=q_B=CV=6C$
When switch is opened, the capacitance of each capacitor becomes $C^{\prime }=KC=3C$
Now potential difference across B is: $V_B=\frac{q_B}{C^{\prime }}=\frac{6C}{3C}=2Volt$