Question

A capacitor of capacitance $ C_{1} $ is charged to a potential $V$ and then connected in parallel to an uncharged capacitor of capacitance $ C_{2} $ .The final potential difference across each capacitor will be

• A. $\frac{ C _{2} V }{ C _{1}+ C _{2}} $
• B. $\frac{C_{1} V}{C_{1}+C_{2}}$
• C. $\left(1+\frac{ C _{2}}{ C _{1}}\right)$
• D. $ \left(1-\frac{ C _{2}}{ C _{1}}\right) V $

Answer 1

The Correct Option is B

To solve the given problem, we need to determine the final potential difference across the capacitors when a charged capacitor is connected to an uncharged one in parallel.

Initial Setup:

• Consider a capacitor with capacitance C_1 charged to a potential difference V.
• It stores a charge Q = C_1 \cdot V.
• An uncharged capacitor with capacitance C_2 is connected in parallel.

Conceptual Understanding:
When two capacitors are connected in parallel, they share the total charge, but the potential difference across each remains the same. The total charge is conserved in the system.

Final Potential Difference Calculation:

1. The total initial charge in the system is Q = C_1 \cdot V.
2. After connection, the total capacitance of the system is (C_1 + C_2).
3. Let V_f be the final potential difference across each capacitor.
4. Using the principle of conservation of charge, we have:
   Q = (C_1 + C_2) \cdot V_f.
5. Substituting Q = C_1 \cdot V in the above equation:
   C_1 \cdot V = (C_1 + C_2) \cdot V_f.
6. Solving for V_f:
   V_f = \frac{C_1 \cdot V}{C_1 + C_2}.

Conclusion:
The final potential difference across each capacitor is \frac{C_1 \cdot V}{C_1 + C_2}. Therefore, the correct option is the second one: \frac{C_1 \cdot V}{C_1 + C_2}.