Question

A parallel plate capacitor is to be designed with a voltage rating $1 \,kV$ using a material of dielectric constant $10$ and dielectric strength $10^{6} V / m ^{-1}$. What minimum area of the plates (in $10^{-3} m ^{2}$ ) is required to have a capacitance of $88.5 \,pF$ ?

Answer 1

For dielectric strength maximum potential difference which can be applied across capacitor is related as
$E_{B}=\frac{V}{d} \Rightarrow d=\frac{V}{E}=\frac{10^{3}}{10^{6}}=10^{-3} m$
We use capacitance of capacitor $C=\frac{\in_{0} \in_{r} A}{d}$
$\Rightarrow A=\frac{C d}{\epsilon_{0} \in_{r}}$
$\Rightarrow A=\frac{88.5 \times 10^{-12} \times 10^{-3}}{8.85 \times 10^{-12} \times 10}=10^{-3} m ^{2}$