What is the analogy between capacitor discharge and radioactive decay?

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AI Thread Summary
The discharge of a capacitor through a resistor is mathematically analogous to radioactive decay, represented by the equations Q = Q_0 e^{-\frac{t}{RC}} and N = N_0 e^{-\lambda t}. In this analogy, A in A = λN corresponds to the rate of change of the quantity, which can be expressed as dN/dt. The negative sign in the current equation I = -Q/(RC) indicates that charge is decreasing, reflecting the direction of current flow. The discussion highlights the importance of understanding how these equations relate to physical processes in both electrical and nuclear contexts.
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Homework Statement



The discharge of a capacitor through a resistor is analogous to radioactive decay.

Write down the equation for capacitor discharge which is analogous to A = \lambda N.

Explain the analogy between A and the subject of your equation.


Homework Equations





The Attempt at a Solution



The only equation I can think of is Q = Q_0 e^{-\frac{t}{RC}} and the similar ones for voltage and current but this can't be the one which is analogous to that equation because it is analogous to N = N_0 e^{-\lambda t}.
 
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You thought of the right equation, you just need to get it in a different form. How do we get the equation N = N_{0}e^{-\lambda{t}}?
 
jdougherty said:
You thought of the right equation, you just need to get it in a different form. How do we get the equation N = N_{0}e^{-\lambda{t}}?

Is the A in A = \lambda N equal to \frac{dN}{dt}?
 
That does seem to be what the question is driving at, yeah.
 
Is the answer I = \frac{Q}{RC}.

I neglected the negative like they did in the question. Why do they get rid of the negative when we talk about activity? Is it because we need it when using differential equations because it is decaying but we leave it when not using the differential equation because we know it is decaying?

What does the negative in I = - \frac{Q}{RC} mean physically? Of course it tells us that charge is decreasing but does it tell us about the direction of current flow?
 
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