Question

A certain parallel-plate capacitor is filled with a dielectric for which κ = 4.61. The area of each plate is 0.0499 m2, and the plates are separated by 2.22 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 246 kN/C. What is the maximum energy that can be stored in the capacitor?

Answer 1

capacitance :-

C = k * (epsilon naught) * A / d

C =4.61 × (8.85 × 10^-12) × (0.0499)/(2.22 × 10^-3)

C = 917.05 pF

Maximum potential difference :-

V = E * d = (246 × 10³) × (2.22 × 10^-3)

V = 546.12 volts

maximum stored energy :-

Energy = ½ * C * V² = (917.05 × 10^-12)(546.12)²/2  

Energy = 136.75 × 10^-6 Joule

Energy = 136.75 μJ