The voltage rating of a parallel plate capaci | vedclass

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(D) Given:Voltage rating $V = 500\,V$Maximum electric field $E_{\max} = 10^6\,V/m$Plate area $A = 10^{-4}\,m^2$Capacitance $C = 15\,pF = 15 	imes 10^

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$8.5$

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(D) Given:Voltage rating $V = 500\,V$Maximum electric field $E_{\max} = 10^6\,V/m$Plate area $A = 10^{-4}\,m^2$Capacitance $C = 15\,pF = 15 	imes 10^{-12}\,F$Permittivity of free space $\epsilon_0 = 8.86 	imes 10^{-12}\,C^2/Nm^2$For a parallel plate capacitor,the maximum voltage $V$ is related to the plate separation $d$ and the maximum electric field $E_{\max}$ by:$V = E_{\max} 	imes d$$d = \frac{V}{E_{\max}} = \frac{500}{10^6} = 5 	imes 10^{-4}\,m$The capacitance of a parallel plate capacitor with a dielectric is given by:$C = \frac{k \epsilon_0 A}{d}$Where $k$ is the dielectric constant.Rearranging for $k$:$k = \frac{C 	imes d}{\epsilon_0 	imes A}$Substituting the values:$k = \frac{15 	imes 10^{-12} 	imes 5 	imes 10^{-4}}{8.86 	imes 10^{-12} 	imes 10^{-4}}$$k = \frac{15 	imes 5}{8.86} = \frac{75}{8.86} \approx 8.465$Rounding to the nearest option,we get $k \approx 8.5$.

The voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of $10^6\,V/m$. The plate area is $10^{-4}\,m^2$. What is the dielectric constant if the capacitance is $15\,pF$? (Given $\epsilon_0 = 8.86 \times 10^{-12}\,C^2/Nm^2$)

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