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: Understanding the Question:
The problem outlines a charge redistribution sequence between two capacitors. Initially, a single capacitor $C_1$ is charged to a potential voltage $V_1$. It is then connected in parallel with an uncharged capacitor $C_2$. We need to derive an expression for the final shared common potential $V_2$.

Step 2: Key Formula or Approach:
According to the

Law of Conservation of Electric Charge, the total charge before sharing must equal the total charge after sharing: $$Q_{\text{initial}} = Q_{\text{final}}$$ 1. Initial charge stored on $C_1$: $Q = C_1 V_1$. 2. When connected in parallel, the two capacitors share the same final voltage $V_2$, and their effective parallel capacitance becomes $C_{\text{eq}} = C_1 + C_2$.
We use the relationship $V = \frac{Q}{C}$ to find the common potential.

Step 3: Detailed Explanation:
Calculate the initial charge present in the circuit before connecting the second capacitor: $$Q_{\text{total}} = C_1 V_1 + C_2(0) = C_1 V_1$$ When the switch is closed and the uncharged capacitor $C_2$ is introduced in parallel, the charge redistributes across the total capacitance: $$C_{\text{total}} = C_1 + C_2$$ Using the basic definition of capacitance, find the new shared common voltage ($V_2$): $$V_2 = \frac{Q_{\text{total}}}{C_{\text{total}}}$$ Substitute our expressions for charge and capacitance into the formula: $$V_2 = \frac{C_1 V_1}{C_1 + C_2}$$ This matches the algebraic structure of option (D).

Step 4: Final Answer:
The shared common potential is $\frac{C_1 V_1}{C_1 + C_2}$, which corresponds to option (D).