A parallel plate capacitor is to be designed | vedclass

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(C) Given:Voltage rating,$V = 1\; kV = 1000\; V$Dielectric constant,$\varepsilon_{r} = 3$Dielectric strength,$E_{max} = 10^{7}\; V/m$Safety limit for

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$19$

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(C) Given:Voltage rating,$V = 1\; kV = 1000\; V$Dielectric constant,$\varepsilon_{r} = 3$Dielectric strength,$E_{max} = 10^{7}\; V/m$Safety limit for electric field,$E = 10\% 	ext{ of } 10^{7} = 10^{6}\; V/m$Capacitance,$C = 50\; pF = 50 	imes 10^{-12}\; F$The distance $d$ between the plates is given by $d = V/E = 1000 / 10^{6} = 10^{-3}\; m$.Using the formula for capacitance $C = \frac{\varepsilon_{0} \varepsilon_{r} A}{d}$,we can find the area $A$:$A = \frac{C \cdot d}{\varepsilon_{0} \cdot \varepsilon_{r}}$Substituting the values:$A = \frac{50 	imes 10^{-12} 	imes 10^{-3}}{8.854 	imes 10^{-12} 	imes 3}$$A = \frac{50 	imes 10^{-15}}{26.562 	imes 10^{-12}} \approx 1.882 	imes 10^{-3}\; m^2$Converting to $cm^2$ $(1\; m^2 = 10^{4}\; cm^2)$:$A \approx 1.882 	imes 10^{-3} 	imes 10^{4} = 18.82\; cm^2 \approx 19\; cm^2$.

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