# Discharge of capacitor question

Gold Member

## 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!