The area of each plate of a parallel plate ca | vedclass

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(C) The dielectric strength of air is $E_{m} = 3 	imes 10^{6} \, V/m$. The maximum potential difference $(V_{m})$ that can be applied across the plat

Vedclass · Accepted Answer

$6000 \, V ; 53 \, nC$

Vedclass · Answer

(C) The dielectric strength of air is $E_{m} = 3 	imes 10^{6} \, V/m$. The maximum potential difference $(V_{m})$ that can be applied across the plates with separation $d = 2 \, mm = 2 	imes 10^{-3} \, m$ is given by: $V_{m} = E_{m} \cdot d = (3 	imes 10^{6} \, V/m) 	imes (2 	imes 10^{-3} \, m) = 6000 \, V$. The capacitance $C$ of the parallel plate capacitor is given by: $C = \frac{\varepsilon_{0} A}{d} = \frac{8.854 	imes 10^{-12} \, F/m 	imes 20 	imes 10^{-4} \, m^2}{2 	imes 10^{-3} \, m} = 8.854 	imes 10^{-12} 	imes 10^{-3} \, F = 8.854 	imes 10^{-12} \, F$. The maximum charge $q_{m}$ on the plates is: $q_{m} = C \cdot V_{m} = (8.854 	imes 10^{-12} \, F) 	imes (6000 \, V) = 53.124 	imes 10^{-9} \, C \approx 53 \, nC$. Thus,the maximum emf is $6000 \, V$ and the charge is $53 \, nC$.

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