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Discharging a capacitor, how variables change over time

  • #1
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Homework Statement



I have a parallel plate capacitor of area A, width d, potential difference V between the plates, and it is disconnected from the source of emf. A wire of resistance R is placed between the plates to allow a current to flow. Draw a sketch to show how the voltage, current and stored energy change over time.

Homework Equations



I = (V/R) exp (-t/RC)

The Attempt at a Solution


The capacitance and resistance are constants I think, as they depend only on the physical characteristics of the setup. So I know that I/V will see an exponential decline over time, tending to zero as t tends to infinity. What I don't know, is how each of them varies individually. I presume the potential difference between the plates will tend to zero, so V will decrease over time (exponentially as well?) and that the current can only flow with a voltage difference. I suspect that they both decline exponentially but can't work out how to demonstrate this.
It might involve I = dQ/dt and V= Ed, but I can't get any further than that.
I guess the energy must decline faster though, as U = 0.5CV^2 and V is declining exponentially.
Any help greatly appreciatec
 

Answers and Replies

  • #2
Redbelly98
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You have correctly found the current I(t).

Since there is just the resistor and capacitor in the circuit, and nothing else, they both have the same current and potential difference. Since you have I, try to think if either the resistor or the capacitor can be used somehow to determine V(t), given the I(t) you have.

(Too bad they chose to call the initial potential difference V, since that is often used for a varying potential difference. So my "V(t)" means the one varying over time, not the initial "V".)
 
  • #3
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Thanks very much.

So I know that for the current across the capacitor:
I(t) = dQ/dt
And that this is also the current across the resistor.

The potential difference across the capacitor is (rearranging the equation given):

V_c(t) = IR exp (t/RC)
V_c(t) = (dQ/dt)R exp (t/RC)

Which is also the potential difference across the resistor. I can use V=Q/C to get

Q/C = (dQ/dt)R exp (t/RC)
Or maybe

I = (Q/CR) exp (-t/RC)
But isn't Q here actually Q(t) as the charge on the plates is a function of time?

This would give me
V = (Q/C) exp (-t/RC)

Meaning that both the current and the potential difference were exponentially declining with time. Is that correct?
 
  • #4
Redbelly98
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Thanks very much.

So I know that for the current across the capacitor:
I(t) = dQ/dt
And that this is also the current across the resistor.

The potential difference across the capacitor is (rearranging the equation given):

V_c(t) = IR exp (t/RC)
V_c(t) = (dQ/dt)R exp (t/RC)
.
.
.
That's fine, but I'll just say it would have gone much easier if you had thought about the resistor instead of the capacitor. How are p.d. and current related to each other for a resistor?


.
.
.
This would give me
V = (Q/C) exp (-t/RC)

Meaning that both the current and the potential difference were exponentially declining with time. Is that correct?
It is correct, however Q and C are not "givens" in the problem statement. Can you write it in terms of the given parameters?
 
  • #5
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Ah, I think I've got it. It comes out as

V(t) = V exp (-t/RC)

Where V is the total charge at t=0. I see what you mean, it's confusing that V is the starting PD and the variable which changes with time...

Thanks very much for your help
 
  • #6
Redbelly98
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You're welcome, glad it worked out.
 

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