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Discharge of capacitor question

  • Thread starter CAF123
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CAF123
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Homework Statement


A charged parallel plate capacitor with plates in the x-y plane and uniform electric field ##\mathbf E = E\hat z## is placed in a uniform magnetic field ##\mathbf B = B \hat x##

i) A resistive wire is connected between the plates in the ##\hat z## direction, so that the capacitor slowly discharges. Show that the total impulse on the wire during the discharge is equal to the momentum orginally stored in the fields.

ii) Instead of discharging the capacitor, slowly reduce the magnetic field to zero. Show again that the resulting total impulse on the plates is equal to the momentum stored in the fields.

Homework Equations


[/B]
momentum stored in the fields ##\mathbf P = \frac{1}{c^2} \int_V d^3 r \mathbf S##

lorentz force.


The Attempt at a Solution


(Working in lorentz heaviside units)[/B]
i) is fine. The momentum stored in the fields is given by above equation, and before the discharge E and B are constant so the momentum is ##(EBAd/c) \hat y## where A is the area of the plates and d is the distance between them.

Then (magnetic) force on wire is $$\mathbf F = \frac{1}{c} \mathbf I \times \mathbf B = \frac{1}{c}Id \hat z \times B \hat z = (dIB/c) \hat y \Rightarrow P_{impulse} = \int F dt = dB/c \int_o^T - \frac{dq}{dt} dt = \dots = \text{result},$$ where t=T is the time when there is zero charge left on the plates.

This gives the correct result but why did the electric field not come into play when it accelerates the charges along the wire?

ii) I am not quite sure how to begin. The changing magnetic field will induce a circular electric field.

Thanks!
 

Answers and Replies

  • #2
205
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Hi. For (i) i think since the charges are accelerated (by the E-field) along the z-direction that doesn't contribute to the wire's momentum (there are forces on the plates but they cancel if the plates are fixed).
For (ii), it seems that by symmetry the E-field created by the changing B-field should only have components along the y axis (think about what B-field two parallel current sheets would create if currents in the sheets ran in the y and –y direction); but in any case we still have to assume the plates are fixed in the z-direction since otherwise the constant E-field would bring them together and the problem would get more complicated...
So the changing B-field would produce: E (z) = (z/c)j ∫dt B(t) , and the total impulse would be: ΔP = qΔE = qB(0)d/c = EBAd/c as advertised
 
  • #3
205
16
Sorry i just realized i shouldn't have spelt out the whole answer, i apologize.. but then i might be wrong! :)
 

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