Question

A parallel plate capacitor is filled with mica (thickness \(1 \times 10^{-3}\) m) and fiber (thickness \(0.5 \times 10^{-3}\) m). Dielectric constants: mica = 8, fiber = 2.5. If fiber breaks at \(6.4 \times 10^6 \, \text{V/m}\), max voltage is:

• A. 3400 V
• B. 5200 V
• C. 2700 V
• D. 4800 V

Hint:

In series dielectric layers, voltages divide as \( V \propto \frac{d}{K} \).

Answer 1

The Correct Option is B

Breakdown field in fiber \( E = 6.4 \times 10^6 \, \text{V/m} \), thickness = \( 0.5 \times 10^{-3} \Rightarrow V = Ed = 6.4 \times 10^6 \cdot 0.5 \times 10^{-3} = 3200 \, \text{V} \)
Since dielectric layer of mica is in series, total voltage = \( V_{\text{mica}} + V_{\text{fiber}} \)
Use field ratio via dielectric constants: \( \frac{V_{\text{mica}}}{V_{\text{fiber}}} = \frac{d_1 K_2}{d_2 K_1} = \frac{1 \cdot 2.5}{0.5 \cdot 8} = \frac{2.5}{4} = 0.625 \)
So, \( V_{\text{mica}} = 0.625 \cdot 3200 = 2000 \)
Total voltage = \( 2000 + 3200 = 5200 \, \text{V} \)