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

Concept: For an isolated capacitor:
•Charge remains constant: \(Q = \text{constant}\)
•Capacitance \(C \propto K\)
•Voltage \(V = \frac{Q}{C} \Rightarrow V \propto \frac{1}{K}\)
•Electric field \(E \propto \frac{V}{d} \Rightarrow E \propto \frac{1}{K}\)

Step 1: Initially \(K_1 = 3\), finally \(K_2 = 9\) \[ \frac{C_2}{C_1} = \frac{K_2}{K_1} = \frac{9}{3} = 3 \]

Step 2: Since capacitor is already charged (isolated): \[ Q_2 = Q_1 = Q_0 \]

Step 3: Voltage relation: \[ V \propto \frac{1}{C} \Rightarrow \frac{V_2}{V_1} = \frac{C_1}{C_2} = \frac{1}{3} \] \[ V_2 = \frac{V_0}{3} \]

Step 4: Electric field: \[ E \propto V \Rightarrow E_2 = \frac{E_0}{3} \] \[ \therefore (Q, V, E) = \left(Q_0, \frac{V_0}{3}, \frac{E_0}{3}\right) \]