Question

A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life τ. Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.

Answer 1

The value of R is 2τC

Explanation:

• A capacitor’s discharge via a resistance R is shown as

• Q = qe - t/CR

• where, Q = Charge left

• C = Capacitance

• q = Initial charge

• R = Resistance

• Energy, E = q2e-2t/CR2C

• Activity, A = A0e - λt

• Here, A0 = Initial activity

• λ = Constant of disintegration  

• Therefore, Energy’s ratio to the activity =

• EA = q2 × e-2t / CR2CA0e-λt

• The terms are not dependent on the time and their coefficients can be compared.

• 2tCR = λt

• ⇒λ = 2CR

• ⇒1τ = 2CR

• ⇒R = 2τC

Answer 2

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