Two capacitors of capacitances 2C and C are joined in parallel and charged to a potential difference V. The battery is then removed and the capacitor of capacitance C is filled completely with a medium of dielectric constant K. What is the new potential difference across the capacitors?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{3V}{K+2} \)
Once the battery is removed, the total charge on the two capacitors stays constant. Capacitances in parallel add, and filling a capacitor completely with a dielectric multiplies its capacitance by \( K \). The common potential difference is the total charge divided by the new total capacitance.
The charge stored before the dielectric is inserted is:
\( \begin{aligned} Q &= (2C+C)\,V \\ &= 3CV \end{aligned} \)
With the capacitor C filled, it becomes \( KC \), and the total capacitance is:
\( \begin{aligned} C' &= 2C+KC \\ &= (K+2)\,C \end{aligned} \)
The new potential difference is:
\( \begin{aligned} V' &= \frac{Q}{C'} \\ &= \frac{3CV}{(K+2)\,C} \\ &= \frac{3V}{K+2} \end{aligned} \)