Question

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 ${10^7}V{m^{ - 1}}$ (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e. without starting to conduct electricity through partial 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 $50pF$ ?

Answer 1

Hint: The electric field between the plates is dependent on potential. We first have to find the maximum electric field that could arise in the capacitor. Then this value can be equated to 10% of dielectric strength.

Complete step-by-step answer:
We know that the capacitance of a parallel plate capacitor is the amount of charge a dielectric can hold when a unit potential difference is given to it.
$C = \dfrac{Q}{V}$ …(1)
We are given that the electric field inside the capacitor should never exceed $10%$ of the dielectric strength.
The dielectric strength ${E_0}$ is given to be ${10^7}V/m$. So the highest Electric Field strength inside should only be $\dfrac{{10}}{{100}}{E_0} = {10^6}V/m$
We are asked to ensure that even for the maximum voltage rated, the Electric field does not cross ${10^6}V/m$
We know the electric field in between the plates of a capacitor is given as :
     …(2)
Here Q is the charge on the capacitor and we may use (1) to find the potential generated due to this charge. Let us replace Q in equation (2) using (1).
     …(3)
Here epsilon is the dielectric constant of the medium. If js the dielectric constant of vacuum and $k$ is the relative permittivity of the dielectric of the capacitor, we can replace as
we know that
So
We are also given that the capacitor should have a capacitance of $50pF$.
So the conditions are that when 1KV of potential difference is given across this $50pF$ capacitor, the Electric field generated inside should be ${10^6}V/m$. So we can substitute these values into our equation to get :
${10^6} = \dfrac{{\left( {50 \times {{10}^{ - 12}}} \right){{10}^3}}}{{A \times 26.55 \times {{10}^{ - 12}}}}$
$A = \dfrac{{50 \times {{10}^3}}}{{{{10}^6} \times 26.55}}$
$A = 1.88 \times {10^{ - 3}}{m^2}$

Note: Using equation (3) to find the area may not be convincing because the capacitance C in the equation is also dependent on A. So if we change the area, shouldn’t we expect the C also to change? But we have to understand that the capacitance is given to have a fixed value. As we change the area, the capacitance could vary. But we can always adjust the distance between the plates and set the capacitance to desired value. Since the electric field between the plates does not depend on the separation, this would not alter the Electric field as well.