The lower plate of a parallel plate capacitor | vedclass

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Question

(B) The electrostatic force of attraction between the plates of a parallel plate capacitor is given by $F = \frac{Q^2}{2A\epsilon_0}$.Since $Q = CV$ a

Vedclass · Accepted Answer

$4.4$

Vedclass · Answer

(B) The electrostatic force of attraction between the plates of a parallel plate capacitor is given by $F = \frac{Q^2}{2A\epsilon_0}$.Since $Q = CV$ and $C = \frac{\epsilon_0 A}{d}$,we have $Q = \frac{\epsilon_0 A V}{d}$.Substituting $Q$ into the force equation: $F = \frac{(\epsilon_0 A V / d)^2}{2A\epsilon_0} = \frac{\epsilon_0 A V^2}{2d^2}$.To maintain balance,this force must be balanced by the weight of the additional mass $m$,so $mg = F$.$m = \frac{\epsilon_0 A V^2}{2d^2g}$.Given: $\epsilon_0 = 8.85 	imes 10^{-12} \, F/m$,$A = 100 \, cm^2 = 10^{-2} \, m^2$,$V = 5000 \, V$,$d = 5 \, mm = 5 	imes 10^{-3} \, m$,$g = 9.8 \, m/s^2$.$m = \frac{8.85 	imes 10^{-12} 	imes 10^{-2} 	imes (5000)^2}{2 	imes (5 	imes 10^{-3})^2 	imes 9.8}$.$m = \frac{8.85 	imes 10^{-14} 	imes 25 	imes 10^6}{2 	imes 25 	imes 10^{-6} 	imes 9.8} = \frac{8.85 	imes 10^{-8}}{19.6 	imes 10^{-6}} \approx 0.004515 \, kg$.$m \approx 4.515 \, g$. Rounding to the nearest option,the correct answer is $4.4 \, g$.

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