A parallel plate capacitor is to be designed with a voltage rating $1 k V$ , using a material of dielectric constant $3$ and dielectric strength about $10^{7} V / m$ . (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionization.) 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 $50 p F$ ?

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Solution

Step 1: Calculate maximum allowed electric field intensity
Given,
Dielectric strength $= 10^{7} V / m$
For safety, the field intensity never exceeds $10 %$ of the dielectric strength.
So, maximum electric field intensity allowed,
$E = 10 %$ of $10^{7} = 10^{6} V / m$

Step 2: Calculate distance between the plates of the capacitor
Formula used: $V = E d$
Given,
Voltage rating on the capacitor
$V = 1 K V = 10^{3} V$
So, distance between the plates of the capacitor,
$d = \frac{V}{E}$

$d = \frac{10^{3}}{10^{6}} = 10^{- 3} m$

Step 3: Calculate minimum required area of the plates
Formula used:
$C = \frac{K ε_{0} A}{d}$

Given,
Dielectric constant of a material, $K = 3$
Capacitance of the parallel plate capacitor, $C = 50 p F = 50 \times 10^{- 12} F$

As, Capacitance
$C = \frac{K ε_{0} A}{d}$

So, area
$A = \frac{C d}{K ε_{0}}$

$A = \frac{50 \times 10^{- 12} \times 10^{- 3}}{3 \times 8.85 \times 10^{- 12}}$

$A = 1.9 \times 10^{- 3} m^{2}$

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