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A parallel plate capacitor is to be designed with a voltage rating $1\; kV$, using a material of dielectric constant $3$ and dielectric strength about $10^7\; Vm^{−1}$. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say $10 \%$ of the dielectric strength. What minimum area of the plates is required to have a capacitance of $50\; pF$?

$\begin{array}{1 1} 300\;cm \\ 30\;cm \\ 19\;cm^2 \\ 60\;cm^2 \end{array} $

1 Answer

Solution :
Potential rating of a parallel plate capacitor, $V = 1\; kV = 1000\; V $
Dielectric constant of a material,$\in =3$
Dielectric strength $= 10^7 V/m$
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, $E = 10\% of 10^7 = 10^6 V/m $
Capacitance of the parallel plate capacitor,
$C = 50 \;pF = 50 \times 10^{−12} F $
Distance between the plates is given by,
$d= \large\frac{V}{E}$
$\quad= \large\frac{1000}{10^6}$
Capacitane is given by the relation
$c= \large\frac{\in_0 \in _rA}{d}$
Where, A = Area of each plate
$\in_0 $ = Permittivity of free space = $8.85 \times 10^{-12}N^{-1}C^{2} m^{-2}$
$A=\large\frac{Cd}{\in_0 \in_r}$
$\qquad= \large\frac{50 \times 10^{-12} \times 10^{-3}}{8.85 \times 10^{-12} \times 3}$
Hence, the area of each plate is about $19\; cm^2. $
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