Question

A parallel plate capacitor is to be constructed which can store $q=10\mu \text{C}$ charge at $V=1000$ Volt . The minimum plate area of the capacitor is required to be $A_1$ when space between the plates has air. If a dielectric of constant $k=3$ is used between the plates, the minimum area required to make such a capacitor is $A_2$. The breakdown field for the dielectric is $8$ times that of air. Then $\frac{A_1}{A_2}$ is:

• A. $12$
• B. $24$
• C. $6$
• D. $3$

Answer 1

The correct option is B $24$

Capacitance of a parallel plate capacitor having dielectric of constant $k$ is
$C=\frac{k{ϵ}_{0}A}{d}$= constant
For $A$ to be minimum, $d$ must be minimum.
The separation between the plates is limited by the breakdown strength of the dielectric.
For air capacitor, $\frac{V}{d_{\text{min}}}=E_{\text{air}}$
[$E_{\text{air}}$] is breakdown field for air.
$\therefore A_1=\frac{CV}{{ϵ}_0{E}_{\text{air}}}$
With dielectric,
$A_2=\frac{CV}{k{ϵ}_0{E}_{\text{diel}}}$
Here [$E_{\text{diel}}$] is breakdown field for dielectric.
$\Rightarrow \frac{A_1}{A_2}=\frac{k{E}_{\text{diel}}}{E_{\text{air}}}=3\times 8=24$