Capacitors moving with constant v

  • #1

I just stumbled apon this question in a past paper that im having trouble understanding, [Broken]

For a) its just straight forward isnt it, C = Aε/d even though they are moving.

b and c im having trouble imagining the direction of the current. If the charges are moving to the right at v, then that would also be the direction the current right? But each plate has a different charge, so the since the top plate is positive the current is to the right with v and as the bottom plate is negative the current is to the left.

Is that the correct way to see it?

Then would you take ampereian loops through the centre of each plate enclosing a section of the plate, so the v vector / I will pass through the loop. One way to explain it is that the area of the loop has elements which point in the y direction, to the right.

I'm getting a little stuck here, i keep thinking what i thought was wrong, but using the ampereian loop doesnt help because the top part of the loop will be outside, so itll get Bl=μIenc then inside itll get negative that because the dl is facing the opposite direction, and the sides of the loop will give zero since their dl is perpendicular to B.

Similarly for the bottom plate as well. And I don't really know what Ienc is, so i cant really say what B is because it doesnt seem right to leave the answer with Ienc.

In the end, adding the field inside gives B=-2uI/l and that would be in the -x direction.

Is that any where near correct?

for c)

q/A (vxB)= q/A vB and that would be in the positive z direction,

that seemed to easy so i must have missed somthing




v= E/B?


magnetic fields do no work, the work was done by electrostatic forces

I think b is the worst so far, but im sure i missed somthing in part c), 5 marks is way to many marks for such little work.

Thanks in advance for any help =]

Homework Statement

Homework Equations

The Attempt at a Solution

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Answers and Replies

  • #2
Hi there,
You've got an interesting question here, and I really hope I could be of some use.
Firstly, since both plates remain at the same location relative to each other, I think you're positively right on a).
For b) have you tried setting a coordinate system at the center of plates, at d/2, and then having a: B(r1)+B(r2), where, depending on the location you'd find non-eliminating components(except at r1=r2=d/2).
Outside them, I think, each would only carry the influence that it creates; In other words, the upper one would produce a field outside itself, dependent solely on it, and the lower likewise and so on(though I could be mistaken).
As for c), It seems about right, though I think the whole deal here is to find the B field, you can use however, B=q(v x r)/r^3(rhat), to determine that.
Does that get you anywhere?
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  • #3
Sweet thanks,

You couldnt use bio-savart for part b) could you because its a sheet of charge and not a line?

For c) i've never seen that formula for B before, with vxr. Do you think the B I would be using for that part of the question would be the B between the plates? Since between the plates the fields due to each plate influence eachother?

Hey do you think my reasoning for e) is ok?

Thanks for helping =]
  • #4
In my view, for b), as you're asked explicitly for the field above/below and in-between, the sheets can be approximated, with fairly good accuracy to a current-carrying wire, and yes, you ought to take the field between them.
Secondly, as for c), B=q(v x r)/r^3(rhat) is a very important derivation of Biot-Savart for moving charges, vital for general considerations via Maxwell's equations. Its form is easily verifiable, and you can read about it here: in greater detail.
As for e), you're right on the mark, so long as the field is in the z direction, and the displacement solely on the x-axis.There will be movement however, on a plain this question doesn't ask us to investigate.
Let me know how it works out,
  • #5
I totally worked it out!

current through the top is to the right, the bottom to the left

put a loop inside the plates so it encloses some current

the 2Bl = Ienc u = u Kl

so for bottom and top B = μσv/2

but outside the fields will cancel eachother because theyre oppositing directions, and inside they will add so Bin = μσv


F/A= KxB = σvB

to balance it has to = F/A for E which will be = σE and E for a single plate is just σ/2ε

then it turns out the force will balance at v = c


thanks for your help =]
  • #6
Well done, but there's really no need to thank me, since you've done everything yourself, and quite so expertly!
Keep up the good work,

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