Capacitor question -- charging time for an RC circuit

[Mausam]

Homework Statement

Hello everyone , this question has been troubling me for weeks,so if anyone could please help me out.i have attached the pic of the question.The answer given is 2 seconds but i am getting a different answer.

Homework Equations

first of all i distributed the charge on the two plates so that there exist no electric field in the conductor.then i calculated the time taken to discharge the charge presant on inside half of the plates But the main thing that's troubling me is how to deal with the charge present on outside of the plate.

The Attempt at a Solution

i have also attached my solution to the question.

 

[BvU]

first of all i distributed the charge on the two plates so that there exist no electric field in the conductor

That's not what the problem statement says. There is a net charge $-Q_0$ on plate A. Closing the switch will increase it to C/V as a function of time. Try to find that function and determine when it crosses 0 Volt.

 

[cnh1995]

That's not what the problem statement says. There is a net charge $-Q_0$ on plate A. Closing the switch will increase it to C/V as a function of time

Should this be C*V?

Try to find that function and determine when it crosses 0 Volt.

Are you saying that the voltage across the capacitor will be zero when the charge on plate A becomes zero?

 

[BvU]

Should this be C*V?

Oh boy, it definitely should !

And with $Q = CV$ I expect Q = 0 when V = 0 ...

 

[cnh1995]

And with Q=CVQ=CVQ = CV I expect Q = 0 when V = 0

Isn't it true only when both the plates of the capacitor have equal and opposite charge? Or am I missing something very fundamental?

This capacitor has only one plate charged initially, which means before connecting it to the battery, some 'net charge' -Q₀ is existing in the circuit. So as charge on A becomes 0, shouldn't charge on B become -Q₀?
If this is true, then I don't think voltage will be zero when Q_A=0.

 

[BvU]

We really want to find the function $Q(t) = C\; V(t)$ .

Charge on B won't go to $-Q_0$ (unless some very specific circumstance applies).

Further hints postponed until OP sets up the equation needed for that.

 

[Mausam]

This capacitor has only one plate charged initially, which means before connecting it to the battery, some 'net charge' -Q0 is existing in the circuit. So as charge on A becomes 0, shouldn't charge on B become -Q0?

The problem states that there exist a net charge -Q₀ initially , now if the switch is open then this charge would get distributed as i have shown in my solution.because in an electrostatic condition there can't be electric field inside the plates(conductors).I am damn sure about it. its the charge outside the plate that's troubling me and according to my solution the final charge on plate B would be -Q_o/2 and that too on the outside.
Thank you for replying, is the answer given correct or my solution has something wrong.

 

[Mausam]

That's not what the problem statement says. There is a net charge $-Q_0$ on plate A. Closing the switch will increase it to C/V as a function of time. Try to find that function and determine when it crosses 0 Volt.

Sir i applied the basic rules for a conductor in an electrostatic condition when the switch was open.And i considered it as net charge only. Could you please tell me at what point am i wrong.According to you what is the charge on respective plates on either sides before closing the switch.

 

[cnh1995]

The problem states that there exist a net charge -Q₀ initially , now if the switch is open then this charge would get distributed as i have shown in my solution.because in an electrostatic condition there can't be electric field inside the plates(conductors).I am damn sure about it. its the charge outside the plate that's troubling me and according to my solution the final charge on plate B would be -Q_o/2 and that too on the outside.
Thank you for replying, is the answer given correct or my solution has something wrong.

Well, I solved it slightly differently (without showing redistribution and just assuming -Q0 on plate A) and I got the same answer as you did. As there is a net charge -Q0 on the capacitor, I believe the sum of the charges on plates A and B at any given instant should be equal to -Q₀ (conservation of charge). This is why I said when Q_A=0, Q_B= -Q₀.

Perhaps I am missing something very fundamental here..
(No wonder I couldn't crack JEE..)

 

[Mausam]

This is why I said when QA=0, QB= -Q0.

oh yeah sorry my bad.But i think that charge redistribution must be considered because charge on the inside half of one plate does not make any sense.Or perhaps we both are missing something crucial.waiting for Bvu's response

 

[gneill]

A schematic doesn't represent the true geometry of a circuit, it only depicts topology. Why not concoct a plausible "real" setup where you can justify charge locations physically? Maybe something like:

Assume that the dimensions of the resistor, switch, and voltage source are tiny compared to the plate size. Assume that their capacitance is much less than that of the plates. I don't show details of how the net negative charge on the left hand side will be distributed, it will depend on the plate thickness and proximity of the plates. Remember that a charged object will be attracted to a neutrally charged body if charge separation takes place.

 

[Mausam]

A schematic doesn't represent the true geometry of a circuit, it only depicts topology. Why not concoct a plausible "real" setup where you can justify charge locations physically? Maybe something like:

View attachment 113251
Assume that the dimensions of the resistor, switch, and voltage source are tiny compared to the plate size. Assume that their capacitance is much less than that of the plates. I don't show details of how the net negative charge on the left hand side will be distributed, it will depend on the plate thickness and proximity of the plates. Remember that a charged object will be attracted to a neutrally charged body if charge separation takes place.

Sir what you posted is completely plausible but our studies(12th grade) we always consider the separation between the plates much smaller than the plates dimensions.And could you please check my solution and tell me where i am wrong.

 

[gneill]

Sir what you posted is completely plausible but our studies(12th grade) we always consider the separation between the plates much smaller than the plates dimensions.

I didn't have enough paper to draw it to scale Consider the plate length (and depth into the page) to be as long as you like to meet the negligible edge effect conditions.

As for your solution, I don't believe that you are permitted to distribute the initial charge on the left plate to the right plate. There is no path for this to occur before the switch is closed.

 

[Mausam]

As for your solution, I don't believe that you are permitted to distribute the initial charge on the left plate to the right plate. There is no path for this to occur before the switch is closed.

why? i have not distributed it from right plate to left, its the induced charge .and i have used basic rules
1)electric field inside conductor is 0 in electrostatic condition
2)conservation of charge
Sir according to you what should be the answer(time taken).is the given answer 2 sec correct?

 

[gneill]

Sir according to you what should be the answer(time taken).is the given answer 2 sec correct?

I find that the given answer of 2 seconds is correct.

 

[Mausam]

I find that the given answer of 2 seconds is correct.

Sir could you please tell me how you got 2 seconds.

 

[gneill]

Sir could you please tell me how you got 2 seconds.

I can't give away the answer. But I can give you a hint as to my thought process.

Consider there to be two populations of charges to deal with (so applying a version of superposition theorem). There's the initial static charge -Q_o and the charge on the capacitor (equal and opposite charges charges on both plates) which is initially zero. The static charge will want to even out, spread equally to both plates. Meanwhile the capacitor charge will be building. At some point the net charge on the left hand plate will be zero. Determine the total amount of charge moved to that point.

 

[Mausam]

Sorry sir i still don't get what you are trying to tell. So according to you half of the static charge would flow through the circuit to plate B ,meanwhile battery does its work of charging the capacitor.so when the charge on plate A is 0 plate B will have -Qo/2 .and the total charge moved to plate A is Qo?

 

[gneill]

Sorry sir i still don't get what you are trying to tell. So according to you half of the static charge would flow through the circuit to plate B ,meanwhile battery does its work of charging the capacitor.so when the charge on plate A is 0 plate B will have -Qo/2 .and the total charge moved to plate A is Qo?

Yes, that's the idea. If you think about it in terms of superposition and conventional current then +Qo/2 flows from plate B to plate A to balance the static charge, and Qo moves from plate B to plate A to charge the capacitor. So a total of (3/2)Qo is moved through the circuit.

The overall circuit (taken as an isolated system) still has a net charge of -Qo, only now it's spread over the entire ensemble rather than gathered on one plate.

 

[gneill]

@Mausam , Looking again at your first post and the charge distribution on the plate surfaces it occurs to me that there may be a less confusing approach than what I've presented above. There I started without the induced charge separation and had it happen as part of the solution process using current flow. That's probably unrealistic and an overly complicated way to account for it.

This time, starting with the charge distribution that you found (which is correct), there is effectively a capacitor already charged to -Q_o/2 (the two facing plate surfaces) holding a potential difference of $V_o = (Q_o/2)/C$ :

Note that the static charge Q_o/2 has already been distributed via the induced charge separation so it exists on the outer surfaces of both plates already, thus it's not necessary to take it into account as a separate current flow as I was doing previously.

Now we can draw an equivalent circuit where the charged capacitor is modeled as an uncharged capacitor in series with a voltage source equal to $V_o$:

You should be able to write an equation for the current with respect to time for this circuit and determine the time it takes for a charge Q_o to flow.

 

[Mausam]

You should be able to write an equation for the current with respect to time for this circuit and determine the time it takes for a charge Qo to flow

I am so sorry for troubling everybody actually i misunderstood the question,it told to find the time taken for net charge on plate to be =0.I thought the charge on both sides of plate A should be zero(which is impossible).i already had the correct solution, but with wrong limits.Thank you all for helping me.

 

[BvU]

Long thread this is becoming ... This is before post # 21 to which I will also reply asap

I find that the given answer of 2 seconds is correct.

@gneill can you point out the error I am making when I agree with @Mausam in post #1 -- except I want to use $Q_0$ instead of $Q_0/2$. So I end up with ${2\ln 2\over \ln 3} = 1.262$ sec ?

It happens to follow the capacitor charging equation with initial condition $Q(0) = -Q_0$ and boundary condition $Q('\infty') = \varepsilon C$ that make A and B in $\ \ Q(t) = Ae^{-t/RC} + B\ \$ assume values of $A = -Q_0 - \varepsilon C$ and $B = \varepsilon C$.

 

[BvU]

I am so sorry for troubling everybody actually i misunderstood the question,it told to find the time taken for net charge on plate to be =0.I thought the charge on both sides of plate A should be zero(which is impossible).i already had the correct solution, but with wrong limits.Thank you all for helping me.

With 1000 V and 1 $\mu$F you'll never get 0.1 C on plate A...

 

[Mausam]

With 1000 V and 1 $\mu$F you'll never get 0.1 C on plate A...

Its not 0.1 ,it is" 0 then a full stop and then i". you confused 1 and I
Anyways i got the correct answer,thank you for your insights.

 

[Mausam]

can you point out the error I am making when I agree with @Mausam in post #1 -- except I want to use Q0Q0Q_0 instead of Q0/2Q0/2Q_0/2. So I end up with 2ln2ln3=1.2622ln⁡2ln⁡3=1.262{2\ln 2\over \ln 3} = 1.262 sec ?

i know you haven't asked me, but i know the mistake so i am replying
if you consider that -Q₀ charge exist on plate A before switch is closed then it violates the fact that in "electrostatic condition" Electric field inside conductor is 0.

 

[gneill]

Long thread this is becoming ... This is before post # 21 to which I will also reply asap

@gneill can you point out the error I am making when I agree with @Mausam in post #1 -- except I want to use $Q_0$ instead of $Q_0/2$. So I end up with ${2\ln 2\over \ln 3} = 1.262$ sec ?

@Mausam had the initial electrostatic charge distribution correct. There's potential energy stored in the circuit thanks to the charge separation that occurs on plate B and this manifests as a pre-existing (before the switch is closed) potential difference between the two plates: $V_o = (Q_o/2)/C$. In this case since Q_o is 1 mC and C = 1 μF, that makes $V_o = 500~V$. The equal charges on the outside surfaces of the plates don't contribute to that potential difference.

If you wish you can think of the circuit as being comprised of a capacitor with an initial voltage of -500 V that will be heading towards +1000 V thanks to the external voltage source. So a ΔV of 1500 V, or a ΔQ of 1500~V / C = 3Q_o/2 going from -Q_o/2 to +Q_o.

 

[BvU]

I'm convinced and impressed. And have actually learned things from this thread ! Thanks both !

 

[cnh1995]

I'm convinced and impressed. And have actually learned things from this thread !

Me too!