Question

A certain parallel-plate capacitor is filled with a dielectric for which κ = 5.0. The area of each plate is 0.027 m2, and the plates are separated by 2.0 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 200 kN/C. What is the maximum energy that can be stored in the capacitor?

Answer 1

Answer:

Energy will be $4.776\times 10^{-5}J$  

Explanation:

We have given that dielectric constant k = 5

Area of the plate $A=0.027m^2$

Distance between the plates $d=2mm=2\times 10^{-3}m$

Electric field E = 200 kN/C

We know that capacitance is given by $C=\frac{K\epsilon _0A}{d}=\frac{5\times 8.85\times 10^{-12}\times 0.027}{2\times 10^{-3}}=0.597\times 10^{-9}F$

We know that electric field is given by $E=\frac{V}{d}$

So $V=Ed=200\times 10^3\times 2\times 10^{-3}=400volt$

So energy will be $E=\frac{1}{2}CV^2=\frac{1}{2}\times 0.597\times 10^{-9}\times 400^2=4.776\times 10^{-5}J$