Reducing Energy in an RC Capacitor: Solving for Discharge Time

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Homework Statement

At what time (in ms) after the discharge begins is the energy stored in the capacitor reduced to half its initial value?

RC=7 ms

Homework Equations

RC=7ms
E=.5CV^2
E=.5Q^2/C
E=.5QV
Q_f = Q_i * e^(-t/RC)

The Attempt at a Solution

I don't even know where to go from here. I know that since the problem involves RC, the equation to solve the problem must have some form of Q_f = Q_i * e^(-t/RC) but I'm stuck.

 

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Homework Statement

At what time (in ms) after the discharge begins is the energy stored in the capacitor reduced to half its initial value?

RC=7 ms

Homework Equations

RC=7ms
E=.5CV^2
E=.5Q^2/C
E=.5QV
Q_f = Q_i * e^(-t/RC)

The Attempt at a Solution

I don't even know where to go from here. I know that since the problem involves RC, the equation to solve the problem must have some form of Q_f = Q_i * e^(-t/RC) but I'm stuck.

Since energy is proportional to the square of charge, by what fraction of the maximum charge should the final charge be to produce half the energy stored with maximum charge?

 

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Since energy is proportional to the square of charge, by what fraction of the maximum charge should the final charge be to produce half the energy stored with maximum charge?

Ah! I see now! When one works the math, the final charge after half the energy is gone is q/2^.5. the initial charge is q. From there it is just plugging into get an the answer 2.43 ms.

Thanks! :D