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A charged capacitor of capacity C is discharged through a resistance R. A radioactive sample decays with an average life J. If the ratio of electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant with time then $R = (x J / C)$ Where x is ______

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answer is 2.

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Detailed Solution

$\frac{E}{A} = \frac{\frac{1}{2 C} {(Q 0 e^{- t / R C})}^{2}}{A_{0} e^{- λ t}}$ For this ratio to be constant, with time; $\frac{2 t}{R C} = λ t$ or $λ = \frac{2}{C R}; \frac{1}{J} = \frac{2}{C R}$

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