Question

A capacitor of unknown capacitance is connected across a battery of \({}'V'\) volt. The charge stored in it is \({}'Q'\) C. When potential across the capacitor is reduced by \({}'V_{1}'\) volt, the charge stored in it becomes \({}'Q_{1}'\) C. The potential \({}'V'\) is

• A. \(\frac{QV_{1}}{Q-Q_{1}}\)
• B. \(\frac{Q}{(Q-Q_{1})}\)
• C. \(\frac{V_{1}}{(Q_{1}-Q)}\)
• D. \(\frac{(Q-Q_{1})}{V_{1}}\)

Hint:

Charge is proportional to voltage, so the drop in charge corresponds to the drop in voltage V1; eliminate the capacitance.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a capacitor \(Q=CV\).

Step 2: Two situations
Initially \(Q=CV\).
After the reduction, \(Q_{1}=C(V-V_{1})\).

Step 3: Calculation
Subtract: \(Q-Q_{1}=CV_{1}\Rightarrow C=\frac{Q-Q_{1}}{V_{1}}\).
\(V=\frac{Q}{C}=\frac{QV_{1}}{Q-Q_{1}}\)

Step 4: Compare the options
This is option (A).

Final Answer:
The potential V equals Q V1 divided by Q minus Q1. \[ \boxed{\text{(A) }\dfrac{QV_{1}}{Q-Q_{1}}} \]