A charged capacitor with capacitance C undergoes discharge through a resistance R . Simultaneously, a radioactive sample decays with an average-life . Determine the value of R such that the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample stays constant over time.

Step-by-step solution

The charge on the capacitor during discharge at time t is given by q(t) = q₀ e^ -t/RC . The electrostatic field energy stored in the capacitor is U(t) = q(t)^2 2C = q₀^2 2C e^ -2t/RC . The activity of a radioactive sample with average life at time t is A(t) = A₀ e^ -t/ . The ratio of the electrostatic field energy to the activity is given by: U(t) A(t) = q₀^2 2C A₀ e^ -t ( 2 RC - 1 ) For this ratio to remain constant over time, the power of the exponential term must be zero. Thus: 2 RC - 1 = 0 2 RC = 1 R = 2 C