Question

A parallel plate capacitor of value 1.77 $\mu$ F is to be designed using a dielectric material (dielectric constant = 200), breakdown strength of $3\times {10}^6$ V/m. In order to make such a capacitor which can withstand a potential difference of 20 V across the plates, the separation between the plates d and area A of the plates respectively are

• A. $6.6\times {10}^{-6}m,{10}^3{m}^2$
• B. $6.6\times {10}^{-5}m,{10}^4{m}^2$
• C. $6.6\times {10}^{-4}m,{10}^5{m}^2$
• D. $6.6\times {10}^{-6}m,{10}^2{m}^2$

Answer 1

Answer: (A)

We knows, electric field is given by
$E=\frac{V}{d}$
$E_{\max }=\frac{V}{d_{\min }}$
$d_{\min }=\frac{V}{E_{\max }}$
Substitute $V=20V$ and $E_{\max }=3\times {10}^{6}V$
$d_{\min }=\frac{20}{3\times {10}^6}=6.6\times {10}^{-6}m$
We knows, the capacitance is given by
$C=\frac{k{\epsilon }_{0}A}{d}$
$\frac{A}{d}=\frac{C}{k{\epsilon }_0}$
$A=\frac{C}{k{\epsilon }_0}$
Putting $C=1.77\mu F=1.77\times {10}^{-6}C,$
$d=6.6\times {10}^{-6},k=200,$
${\epsilon }_0=8.85\times {10}^{-12}{C}^2{N}^{-1}{m}^2$
$A=\frac{1.77\times {10}^{-6}\times 6.6\times {10}^{-6}}{200\times 8.85\times {10}^{-12}}={10}^3{m}^2$