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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 $10^{7}\; V\,m ^{-1} .$ (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 (in $cm^2$) of the plates is required to have a capacitance of $50\; pF ?$
$26$
$8$
$19$
$33$
Solution
Potential rating of a parallel plate capacitor, $V =1 \,kV =1000 \,V$
Dielectric constant of a material, $\varepsilon_{r}=3$
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, $d=\frac{V}{E}$
$=\frac{1000}{10^{6}}=10^{-3} \,m$
Capacitance is given by the relation, $C=\frac{\epsilon_{0} \epsilon_{,} A}{d}$
Where,
$A=$ Area of each plate
$\epsilon_{0}=$ Permittivity of free space $=8.85 \times 10^{-12} \,N ^{-1} \,C ^{2} \,m ^{-2}$
$\therefore A =\frac{C d}{\epsilon_{0} \in}$
$=\frac{50 \times 10^{-12} \times 10^{-3}}{8.85 \times 10^{-12} \times 3} \approx 19 \,cm ^{2}$
Hence, the area of each plate is about $19\; cm ^{2}$.
