Dis/Charging capacitor in time varying electric field

[Apteronotus]

I'm interested in calculating the capacitive current when the electric field across a capacitor is given by a random function. The randomness of the electric field deems the usual $$I_c=C\frac{dV}{dt}$$ useless; since the derivative of a random function $$V$$ cannot be calculated.

To approach the problem in another way, I'd like to calculate the charge $$q(t)$$ on a capacitor in an RC circuit when the emf is given by a step function:
$$V(t)=V_1$$ for $$t\in(0,t_1)$$
$$V(t)=V_2$$ for $$t\in(t_1,t_2)$$
and
$$V(t)=V_3$$ for $$t\in(t_2,t_3)$$
where
$$V_1<V_3<V_2$$
I've calculated the charge $$q(t)$$ for $$t\in(t_1,t_2)$$ to be
$$q(t)=q_1+C(V_2-V1)(1-e^{\frac{t_1-t}{RC}})$$
but am having a difficult time deriving the charge in the interval $$(t_2,t_3)$$.

If anyone can help me with either of the problems I'd much appreciate it.

Thanks.

 

[berkeman]

I'm interested in calculating the capacitive current when the electric field across a capacitor is given by a random function. The randomness of the electric field deems the usual $$I_c=C\frac{dV}{dt}$$ useless; since the derivative of a random function $$V$$ cannot be calculated.

To approach the problem in another way, I'd like to calculate the charge $$q(t)$$ on a capacitor in an RC circuit when the emf is given by a step function:
$$V(t)=V_1$$ for $$t\in(0,t_1)$$
$$V(t)=V_2$$ for $$t\in(t_1,t_2)$$
and
$$V(t)=V_3$$ for $$t\in(t_2,t_3)$$
where
$$V_1<V_3<V_2$$
I've calculated the charge $$q(t)$$ for $$t\in(t_1,t_2)$$ to be
$$q(t)=q_1+C(V_2-V1)(1-e^{\frac{t_1-t}{RC}})$$
but am having a difficult time deriving the charge in the interval $$(t_2,t_3)$$.

If anyone can help me with either of the problems I'd much appreciate it.

Thanks.

I'm not understanding the setup yet. The electric field between the plates is determined by the plate voltage difference. Are you suggesting somehow to change that?

 

[Apteronotus]

Hi Berkeman,

Lets simply suppose we have an RC circuit connected to a voltage device (battery).
For the first time interval $$(0,t_1)$$ the battery voltage is $$V_1$$ for the second interval $$(t_1,t_2)$$ the voltage is $$V_2$$ and so on...

I want to measure the the amount of charge that would accumulate on the capacitor as a function of time.

For the first time interval, we simply have
$$q(t)=CV_1(1-e^{-\frac{t}{RC}})$$
for the second time interval we'd get
$$q(t)=CV_1(1-e^{-\frac{t_1}{RC}}) + C(V_2-V_1)(1-e^{-\frac{t_1-t}{RC}})$$

now considering that $$V_3<V_2$$ the capacitor need to discharge slightly.
So what is q(t) in the third time interval?

Thanks.

 

[berkeman]

Hi Berkeman,

Lets simply suppose we have an RC circuit connected to a voltage device (battery).
For the first time interval $$(0,t_1)$$ the battery voltage is $$V_1$$ for the second interval $$(t_1,t_2)$$ the voltage is $$V_2$$ and so on...

I want to measure the the amount of charge that would accumulate on the capacitor as a function of time.

For the first time interval, we simply have
$$q(t)=CV_1(1-e^{-\frac{t}{RC}})$$
for the second time interval we'd get
$$q(t)=CV_1(1-e^{-\frac{t_1}{RC}}) + C(V_2-V_1)(1-e^{-\frac{t_1-t}{RC}})$$

now considering that $$V_3<V_2$$ the capacitor need to discharge slightly.
So what is q(t) in the third time interval?

Thanks.

It looks like you are on the right track. Just do what you did fror the 2nd equation. Fix the 2nd term at t2 to make hold whatever it was at at the end of t2, and write a similar 3rd equation with the voltage difference with respect to that fixed 2nd term. The capacitor sort of has "memory" in that it is storing some amount of charge, based on its past histopry and the Vn voltages and tn time intervals.