Question

A certain parallel-plate capacitor is filled with a dielectric for which κ = 4.14. The area of each plate is 0.0360 m2, and the plates are separated by 1.68 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 220 kN/C. What is the maximum energy that can be stored in the capacitor?

Answer 1

Answer:

Maximum energy of the capacitor, $E=5.36\times 10^{-5}\ J$    

Explanation:

It is given that,

Dielectric constant of the parallel plate capacitor, k = 4.14

Area of cross section of each plate, $A=0.0360\ m^2$

Separation between the plates, d = 1.68 mm = 0.00168 m

The electric field between the plates, $E=220\ kN/C=220\times 10^3\ N/C$

Let E is the maximum energy that can be stored in the capacitor. Its formula is given by :

$E=\dfrac{1}{2}CV^2$

C is the capacitance of the capacitor,      

$C=\dfrac{kA\epsilon_o}{d}$

$C=\dfrac{4.14\times 0.0360\times 8.85\times 10^{-12}}{0.00168}$

$C=7.85\times 10^{-10}\ F$

V is the potential difference,

$V=E\times d$

$V=220\times 10^3\times 0.00168$

V = 369.6 volts

$E=\dfrac{1}{2}CV^2$

$E=\dfrac{1}{2}\times 7.85\times 10^{-10}\times (369.6)^2$  

$E=5.36\times 10^{-5}\ J$    

So, the maximum energy that can be stored in the capacitor is$5.36\times 10^{-5}\ J$. Hence, this is the required solution.