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RC capacitor and energy

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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.
 

Answers and Replies

  • #2
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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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  • #3
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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 in to get an the answer 2.43 ms.

Thanks!! :D
 

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