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(C) The electric field $E$ between the plates is proportional to the potential difference $V$ across the capacitor $(E = V/d)$.Given that the electric

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(C) The electric field $E$ between the plates is proportional to the potential difference $V$ across the capacitor $(E = V/d)$.Given that the electric field drops to one-third of its initial value,the potential difference also drops to one-third: $V = V_0 / 3$.The discharging equation for a capacitor is $V = V_0 e^{-t/	au}$,where $	au = RC$ is the time constant.Substituting the given values: $V_0 / 3 = V_0 e^{-t/	au} \Rightarrow 1/3 = e^{-t/	au}$.Taking the natural logarithm on both sides: $\ln(3) = t/	au$.Given $\ln(3) = 1.1$,$t = 6.6 	imes 10^{-6} \ s$,and $C = 1.5 	imes 10^{-6} \ F$.$1.1 = (6.6 	imes 10^{-6}) / (R 	imes 1.5 	imes 10^{-6})$.$1.1 = 6.6 / (1.5 	imes R)$.$R = 6.6 / (1.5 	imes 1.1) = 6.6 / 1.65 = 4 \ \Omega$.

The electric field between the two parallel plates of a capacitor of $1.5 \mu F$ capacitance drops to one third of its initial value in $6.6 \mu s$ when the plates are connected by a thin wire. The resistance of this wire is . . . . . . . $\Omega$. (Given,$\log_{e} 3 = 1.1$)

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