A certain parallel-plate capacitor is filled with a dielectric for which κ = 5.85. The area of each plate is 0.0817 m2, and the plates are separated by 1.96 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 222 kN/C. What is the maximum energy that can be stored in the capacitor?

Answer: 204.29 * 10^-6 J

Explanation: In order to explain this question we have to consider to following expression for the potencial and electric field in a capacitor, which is given by;

E*d=V where d is the separation between plates and E and V are the electric field and the voltage difference in the plates.

For the maximun electric field, V max es equal

Vmax=Emax*d = 222*10^3 * 1.96*10^-3=435.12 V

We also know that potential energy stores by a capacitor with a constant dielectric κ is given by:

E=0.5*C*V^2, where C is the capacity and V the voltage.

For two plates capacitor C is equalk to:

C = k*ε*A/d where A and d are the area and the separation in the capacitor. εo is a constant equal 8.85*10^-12 F/m

Then;

E=[ (0.5*5.85*8.85*10^-12*0.0817) / (1.96*10^-3) ] * (435.12) ^2 = 204.29 * 10^-6 J