Griffiths 8.5: Impulse and Momentum parallel plate capacitor

In summary: 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.
  • #1
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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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 

1. What is the equation for calculating impulse in a parallel plate capacitor?

The equation for calculating impulse in a parallel plate capacitor is given by I = QV, where I is the impulse, Q is the charge, and V is the potential difference between the plates.

2. How is momentum related to the motion of charges in a parallel plate capacitor?

Momentum is related to the motion of charges in a parallel plate capacitor through the equation p = mv, where p is the momentum, m is the mass of the charges, and v is the velocity at which they are moving.

3. Can the impulse and momentum of a parallel plate capacitor be changed?

Yes, the impulse and momentum of a parallel plate capacitor can be changed by altering the charge, potential difference, or mass of the charges within the capacitor.

4. How does the distance between the plates of a parallel plate capacitor affect the impulse and momentum?

The distance between the plates of a parallel plate capacitor does not directly affect the impulse and momentum, but it can indirectly affect them by changing the electric field strength and therefore the force on the charges, which in turn affects their acceleration and momentum.

5. What is the significance of impulse and momentum in a parallel plate capacitor?

Impulse and momentum are important concepts in a parallel plate capacitor because they allow us to understand the motion of charges within the capacitor and how it is affected by factors such as charge, potential difference, and distance between plates. They also have practical applications in areas such as circuit design and energy storage.

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