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**Summary::**Griffiths problem 8.5

Problem 8.5 of Griffiths (in attachment)

I already solved part (a), and found the momentum in the fields to be $$\textbf{p}=Ad\mu_0 \sigma^2 v \hat{\textbf{y}}$$

In part (b), I am asked to find the total impulse imparted on the plates if the top plate starts moving downwards.

Since the electrostatic attraction cancels out, I find that the only components of the impulse are:

1) the magnetic repulsion of top plate by bottom plate

2) the magnetic repulsion of bottom plate by top plate

3) as the top plate moves down, the magnetic field above drops to zero inducing an electric field which acts on the bottom plate

However, doing so I find that the first two components cancel out, and that therefore the only component to the impulse is that of the induced electric field, which is half the momentum stored in the fields.

I have seen some solutions and they don't take into account the magnetic repulsion of bottom plate by top plate, why so?

**[Moderator's note: Moved from a technical forum and thus no template.]**

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