Question

An uncharged parallel plate capacitor with dielectric constant \(K\) is connected to a similar air-cored capacitor which is charged to a potential \(V_0\). If the common potential is \(V\), then the value of \(K\) is

• A. \(\frac{V_0}{V} - 1\)
• B. \(\frac{V_0}{V} + 1\)
• C. \(\frac{V}{V_0} - 1\)
• D. \(\frac{V}{V_0} + 1\)

Hint:

Use charge conservation and final potential formula: \(V = \frac{Q}{C_1 + C_2}\)

Answer 1

The Correct Option is A

Let: - Capacitance of both capacitors (without dielectric) be \(C\)
- One capacitor has dielectric → capacitance becomes \(KC\)
- The air capacitor is initially charged to potential \(V_0\), so its charge is: \[ Q = CV_0 \] Now they are connected. The total charge is conserved, and the final potential is \(V\).
Total capacitance after connection = \(C + KC\)
Final common potential: \[ V = \frac{\text{Total Charge}}{\text{Total Capacitance}} = \frac{CV_0}{C + KC} \Rightarrow V = \frac{V_0}{1 + K} \Rightarrow 1 + K = \frac{V_0}{V} \Rightarrow K = \frac{V_0}{V} - 1 \]

Answer 2

Step 1: Understand the setup
- Two capacitors of same size: one with dielectric constant \(K\), initially uncharged.
- Another air-cored capacitor charged to potential \(V_0\).
- After connecting, common potential becomes \(V\).

Step 2: Recall capacitance of capacitors
- Capacitance of air-cored capacitor: \(C_0\)
- Capacitance of dielectric capacitor: \(C = K C_0\)

Step 3: Use charge conservation
Initial charge on charged capacitor:
\[ Q_0 = C_0 V_0 \]
Total charge after connection is shared, so:
\[ Q_\text{total} = Q_0 = (C_0 + C) V = (C_0 + K C_0) V = C_0 (1 + K) V \]

Step 4: Equate charges and solve for \(K\)
\[ C_0 V_0 = C_0 (1 + K) V \Rightarrow V_0 = (1 + K) V \Rightarrow K = \frac{V_0}{V} - 1 \]

Step 5: Final answer
The dielectric constant is \(K = \frac{V_0}{V} - 1\).