Applying Gauss' Law for Calculating Work Req'd to Move Charge

[lawsonfurther]

Homework Statement

A capacitor has two square plates that are d apart. Each plate is L×L. The capacitor is initially uncharged.
(a)Calculate the work required to move q of charge from one plate to the other.
(b)Calculate the work required to move an additional q of charge from one plate to the other.

Homework Equations

Gauss's law
definition of work

The Attempt at a Solution

I don't quite understand how to apply the Gauss's law to get the electric field.
My attempt was that W=Fd=qEd. And to find the electric field strength E, we need to imagine a Gaussian surface enclosing the plate from where you want to remove the charge, which will be a pill box in this case. So when I am trying to find out the area where the field goes through, should I count only one side of the pill box(either top or bottom) or both sides of it?
From the correct answer provided as W_(a)=q^2d/(2L^2(E0)), where E0 is the epsilon in Gauss's law, it seems like I should count both sides of the imaginary pill box. But I am not quite convinced of it.
Can anyone explain it to me?

 

[haruspex]

the area where the field goes through, should I count only one side of the pill box(either top or bottom) or both sides of it?

For two equally and oppositely charged plates, placed close together, how do you think the field lines are arranged?

 

[lawsonfurther]

For two equally and oppositely charged plates, placed close together, how do you think the field lines are arranged?

The field lines will point from the positively charged plate to the negatively charged plated and they will be perpendicular to the surface of the plates, if ignoring the edge effect.

 

[haruspex]

The field lines will point from the positively charged plate to the negatively charged plated and they will be perpendicular to the surface of the plates, if ignoring the edge effect.

Right, so all the field lines go through one side of the pill box only. There is nothing to count on the other side.
Or, looking at it another way, in the region outside the plates the two fields cancel.

 

[lawsonfurther]

Right, so all the field lines go through one side of the pill box only. There is nothing to count on the other side.
Or, looking at it another way, in the region outside the plates the two fields cancel.

Okay. But in question (a) when those charges are on their way from one plate to the other plate which is initially neutral, the other plate shouldn't have any charge until those charges reach it. So in other words, shouldn't there be no electric field coming from the destination plate until the whole process is done?

 

[haruspex]

no electric field coming from the destination plate until the whole process is done?

No. Transfer a small amount of charge at a time. The potential increases all the while. It is an integral.

 

[lawsonfurther]

No. Transfer a small amount of charge at a time. The potential increases all the while. It is an integral.

Well, I still don't get it. If I count only one side of the pill box, then E(L^2)=q/(epsilon) which means E=q/(E0L^2) with E0 being epsilon. Then if I use W=qEd, then the work I get will be q^2d/(E0L^2). Am I finding the right thing?

 

[haruspex]

Well, I still don't get it. If I count only one side of the pill box, then E(L^2)=q/(epsilon) which means E=q/(E0L^2) with E0 being epsilon. Then if I use W=qEd, then the work I get will be q^2d/(E0L^2). Am I finding the right thing?

The charge increases from 0 to q.
Suppose at some point we have moved a charge Q across. What is the field now? What work is required to move the next instalment ΔQ across?
(It is unfortunate they used d for the plate separation. Can we agree to call it D instead? Otherwise there will be confusion with calculus notation.)

 

[lawsonfurther]

Well, if we have done moving the charge Q to the other plate, the field between two plates should have been q/(E0L^2) as I found above. But after that, we need to remove another ΔQ again from the original plate. So shouldn't the field be altered during the process of transferring that amount of ΔQ?
Or are you saying that for the first part of the question, the charge I removed will create a field that will partially cancel out the field created by the original plate so that the net electric field outside the original plate (not in between the plates) should be zero?

 

[haruspex]

if we have done moving the charge Q to the other plate, the field between two plates should have been q/(E0L^2)

The question specifies "q" as the total charge moved. I introduced "Q" to represent the charge moved so far at some stage in the process. So Q starts at 0 and increases to q.

shouldn't the field be altered during the process of transferring that amount of ΔQ?

Sure, but ΔQ is small, so the increase in the field has little effect on the work required to move that ΔQ. This is how integration works.
So how much work is required to move the ΔQ?