Question

Suppose that a parallel-plate capacitor has a dielectric that breaks down if the electric field exceeds K V/m. Thus, the maximum voltage rating of the capacitor is Vmax=Kd, where d is the separation between the plates. In working problem P3.33, we find that the maximum energy that can be stored is wmax=12ϵrϵ0K2(Vol) in which Vol is the volume of the dielectric. Given that K = 32 × 105V/m and that ϵr = 1 (the approximate values for air), find the dimensions of a parallel-plate capacitor having square plates if it is desired to store 1 mJ at a voltage of 2000 V in the least possible volume.

Answer 1

Answer:

The least possible volume is $2.206*10^{-5} m^3$

Explanation:

We are asked by the least volume to store 1 mJ at 2000 V.

So, we can use the given formula for $W_{max}$ and solve for Vol.

$W_{max} = \frac{1}{2}\epsilon_r \epsilon_0 K^2 Vol\\Vol= 2W_{max}/(\epsilon_r \epsilon_0 K^2)$

replacing the values given in the problem and the permittivity of space $\epsilon_0$ which is $8.854*10^{-12}F/m$ we obtain Vol.

$Vol= 2*0.001 J/(1*8.854*10^{-12}F/m*(32*10^5V/m)^2)\\Vol= 0.002 J/( 90.66496F*V^2/m^3)\\Vol = 2.206*10^{-5} m^3$

Note that $FV^2= J$ in the above solution

Additional:

Taking into account that the volume of dieletric will be the area of plates (A) times the separation between plates (d).

$Vol=A*d$

You also can calculate A and d

d is calculated assuming that the $V_{max}$ is 2000 V and using given equation for $V_{max}$:

$d= V_{max}/K\\d= 2000V/(32*10^5V/m)\\d=6.25*10^-4 m$

and A is calculated dividing Vol by d

$A= Vol/d=2.206*10^{-5} m^3/6.25*10^-4 m =0.0353 m^2$