Griffiths 8.5: Impulse and Momentum parallel plate capacitor

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Homework Help Overview

The discussion revolves around Griffiths problem 8.5, which involves analyzing the impulse and momentum in a parallel plate capacitor as one plate moves downwards. The original poster has solved part (a) and is now focused on part (b), which requires finding the total impulse on the plates due to the movement and the associated electromagnetic effects.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster discusses the components of impulse, including magnetic repulsion and the induced electric field. They question why some solutions do not consider the magnetic repulsion between the plates. Other participants clarify that forces acting perpendicularly to the direction of momentum do not contribute to the impulse in that direction.

Discussion Status

The discussion is ongoing, with participants exploring the implications of electromagnetic momentum and the directionality of forces. There is a recognition of the need to focus on the impulse associated with the loss of field momentum in the specified direction.

Contextual Notes

Participants are navigating assumptions about the contributions of various forces and the specific setup of the problem, particularly regarding the direction of impulse and momentum in the context of the capacitor's operation.

KDPhysics
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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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KDPhysics said:
I have seen some solutions and they don't take into account the magnetic repulsion of bottom plate by top plate, why so?
The initial electromagnetic momentum in the fields between the plates is in the ##\hat y## direction. Therefore, the impulse associated with the loss of this field momentum must be associated with forces acting parallel to ##\hat y##. Since any attractive or repulsive forces between the plates act perpendicularly to ##\hat y##, these forces do not contribute any impulse in the ##\hat y##-direction.
 
TSny said:
The initial electromagnetic momentum in the fields between the plates is in the ##\hat y## direction. Therefore, the impulse associated with the loss of this field momentum must be associated with forces acting parallel to ##\hat y##. Since any attractive or repulsive forces between the plates act perpendicularly to ##\hat y##, these forces do not contribute any impulse in the ##\hat y##-direction.
So then I should not take into consideration the magnetic repulsion?
 
KDPhysics said:
So then I should not take into consideration the magnetic repulsion?
I believe that's right. I think the purpose of the problem is to understand what's going on with momentum and impulse in the ##\hat y## direction.
 

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