A parallel plate capacitor is to be designed, | vedclass

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(D) The dielectric strength $E$ is the maximum electric field the dielectric can withstand. The separation $d$ between the plates is given by $d = \fr

Vedclass · Accepted Answer

$21.7 	imes 10^{-6} \; m^2$

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(D) The dielectric strength $E$ is the maximum electric field the dielectric can withstand. The separation $d$ between the plates is given by $d = \frac{V}{E}$.Substituting the given values,$V = 12 	imes 10^3 \; V$ and $E = 10^9 \; Vm^{-1}$,we get $d = \frac{12 	imes 10^3}{10^9} = 12 	imes 10^{-6} \; m$.The capacitance of a parallel plate capacitor is $C = \frac{K \varepsilon_0 A}{d}$.Rearranging for area $A$,we get $A = \frac{C d}{K \varepsilon_0} = \frac{C V}{K E \varepsilon_0}$.Substituting the values $C = 80 	imes 10^{-12} \; F$,$V = 12 	imes 10^3 \; V$,$K = 5$,$E = 10^9 \; Vm^{-1}$,and $\varepsilon_0 = 8.85 	imes 10^{-12} \; Fm^{-1}$:$A = \frac{80 	imes 10^{-12} 	imes 12 	imes 10^3}{5 	imes 10^9 	imes 8.85 	imes 10^{-12}} = \frac{960 	imes 10^{-9}}{44.25 	imes 10^{-3}} \approx 21.69 	imes 10^{-6} \; m^2$.Thus,the minimum area is approximately $21.7 	imes 10^{-6} \; m^2$.

$A$ parallel plate capacitor is to be designed,using a dielectric of dielectric constant $5$,so as to have a dielectric strength of $10^9 \; Vm^{-1}$. If the voltage rating of the capacitor is $12 \; kV$,the minimum area of each plate required to have a capacitance of $80 \; pF$ is

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