Current in Discharing Capacitor

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AI Thread Summary
The discussion focuses on how to express the current in a discharging capacitor, starting from the relationship I = dQ/dt. The user presents the charge equation Qd = Q0e^(-t/RC) and attempts to differentiate it to find the current. There is confusion regarding the correct derivative, with a suggestion that the negative sign may be overlooked. The conversation highlights the importance of understanding the sign of current, as it can be represented as positive or negative depending on the direction. Clarifying the derivative and its implications for current direction is essential for accurate understanding.
05holtel
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Homework Statement



How do you express the current in a discharging capacitor.

I know that I = dQ/dt,

Qd = Q0e^(-t/RC)
I = d/dt (Q0e^(-t/RC))
= (Q/RC)e^(-t/RC)

I am confused because taking the derivative of Q0e^(-t/RC)) I think should be
(-Q/RC)e^(-t/RC) because the derivative of (-t/RC) is (-1/RC)

Which solution is correct
Thanks in advance
 
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Don't quote me on this but they may just be taking the absolute value for simplicity, since current can be taken as positive in one direction or negative to point in the other direction.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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