A capacitor of capacitance \( C_1 \) is charged upto V volt and then connected to an uncharged capacitor of capacity \( C_2 \). The final potential difference across each will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \frac{C_1 V}{\left(C_1+C_2\right)} \)
When the two capacitors are joined, charge flows until they reach a common potential. They are then in parallel, and the total charge is conserved.
\( \text{Charge before} = C_1 V \)
\( \text{Charge after} = (C_1 + C_2)\,V' \)
\( \displaystyle C_1 V = (C_1 + C_2)\,V' \;\Rightarrow\; V' = \frac{C_1 V}{C_1 + C_2} \)