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(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.In time $t_{1}$,the energy reduces to half,so $U(t_{1}) = \frac{U_{0}}{2}$.This

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$1/6$

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(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.In time $t_{1}$,the energy reduces to half,so $U(t_{1}) = \frac{U_{0}}{2}$.This implies $\frac{q(t_{1})^2}{2C} = \frac{1}{2} \frac{Q_{0}^2}{2C}$,which simplifies to $q(t_{1}) = \frac{Q_{0}}{\sqrt{2}}$.Using the discharging equation $q(t) = Q_{0} e^{-t/RC}$,we have $\frac{Q_{0}}{\sqrt{2}} = Q_{0} e^{-t_{1}/RC}$,so $e^{-t_{1}/RC} = 2^{-1/2}$.Taking the natural logarithm,$\frac{t_{1}}{RC} = \frac{1}{2} \ln(2)$.In time $t_{2}$,the charge reduces to one-eighth,so $q(t_{2}) = \frac{Q_{0}}{8}$.Using the discharging equation,$\frac{Q_{0}}{8} = Q_{0} e^{-t_{2}/RC}$,so $e^{-t_{2}/RC} = 2^{-3}$.Taking the natural logarithm,$\frac{t_{2}}{RC} = 3 \ln(2)$.Now,the ratio $\frac{t_{1}}{t_{2}} = \frac{\frac{1}{2} \ln(2)}{3 \ln(2)} = \frac{1}{6}$.

$A$ capacitor is discharging through a resistor $R$. Consider in time $t_{1}$,the energy stored in the capacitor reduces to half of its initial value and in time $t_{2}$,the charge stored reduces to one-eighth of its initial value. The ratio $t_{1} / t_{2}$ will be ................

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