Question

A condenser of capacity $C_1$ is charged to a potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is then connected in parallel with $C_1$. The resultant common potential $V_2$ is

• A. $\frac{V_1 C_2}{C_1}$
• B. $\frac{C_2}{C_1 + C_2}$
• C. $\frac{C_1 V_1}{C_2}$
• D. $\frac{C_1 V_1}{C_1 + C_2}$

Hint:

You can verify the formula quickly using limiting case analysis! If the second capacitor $C_2$ is extremely tiny ($C_2 \to 0$), the formula simplifies to $\frac{C_1 V_1}{C_1} = V_1$, which makes sense because the voltage shouldn't drop if no capacity was added.

Answer 1

The Correct Option is D

Step 1: The story.
A capacitor $C_1$ is charged to voltage $V_1$, then unplugged. An empty capacitor $C_2$ is joined in parallel with it. We want the common voltage $V_2$ they settle to.
Step 2: Charge is conserved.
No charge can leak away, so the total charge before joining equals the total after joining.
Step 3: Charge before joining.
Only $C_1$ has charge, and $C_2$ is empty: \[ Q_{\text{total}} = C_1 V_1 + 0 = C_1 V_1 \]
Step 4: Capacitance after joining.
In parallel the capacitances add: \[ C_{\text{total}} = C_1 + C_2 \]
Step 5: Use the charge formula.
The shared voltage is total charge over total capacitance: \[ V_2 = \frac{Q_{\text{total}}}{C_{\text{total}}} \]
Step 6: Substitute and finish.
\[ V_2 = \frac{C_1 V_1}{C_1 + C_2} \] This is option (4). \[ \boxed{V_2 = \frac{C_1 V_1}{C_1 + C_2}} \]