Question

A charged capacitor when filled with a dielectric \(K = 3\) has charge \(Q_0\), voltage \(V_0\) and field \(E_0\). If the dielectric is replaced with another one having \(K = 9\), the new values of charge, voltage and field will be respectively:

• A. \(3Q_0, 3V_0, 3E_0\)
• B. \(Q_0, 3V_0, 3E_0\)
• C. \(Q_0, \frac{V_0}{3}, 3E_0\)
• D. \(Q_0, \frac{V_0}{3}, \frac{E_0}{3}\)

Hint:

If capacitor is isolated → charge constant; if connected to battery → voltage constant. Always check this!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
If a charged capacitor is isolated (not connected to a battery), its charge \(Q\) remains constant even if a dielectric is inserted or changed. However, its capacitance, voltage, and internal electric field change depending on the dielectric constant \(K\).
Step 2: Key Formula or Approach:
1. Capacitance \(C = K C_{\text{air}}\).
2. Potential difference \(V = \frac{Q}{C} = \frac{Q}{K C_{\text{air}}}\).
3. Electric field \(E = \frac{V}{d} = \frac{Q}{K C_{\text{air}} d}\).
Step 3: Detailed Explanation:
Let the air capacitance be \(C_{a}\).
Initial state (\(K_{1} = 3\)):
\(Q_{\text{initial}} = Q_{0}\).
\(V_{0} = \frac{Q_{0}}{3 C_{a}}\).
\(E_{0} = \frac{Q_{0}}{3 C_{a} d}\).
New state (\(K_{2} = 9\)):
1. Charge remains constant: \(Q_{\text{new}} = Q_{0}\).
2. New potential \(V_{\text{new}} = \frac{Q_{0}}{9 C_{a}}\).
By comparison: \(V_{\text{new}} = \frac{1}{3} \left( \frac{Q_{0}}{3 C_{a}} \right) = \frac{V_{0}}{3}\).
3. New field \(E_{\text{new}} = \frac{V_{\text{new}}}{d} = \frac{V_{0}/3}{d} = \frac{E_{0}}{3}\).
Step 4: Final Answer:
The new values are \(Q_{0}, \frac{V_{0}}{3}, \frac{E_{0}}{3}\).