What is the force on a point charge in Griffiths problem 8.4?

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SUMMARY

The discussion focuses on Griffiths problem 8.4, which involves calculating the force between two equal point charges, q, separated by a distance of 2a, using Maxwell's stress tensor. The relevant equation for the stress tensor is T_{ij} = \epsilon_{0} (E_{i}E_{j} - \frac{1}{2} \delta_{ij} E^2) + \frac{1}{\mu_{0}} (B_{i}B_{j} - \frac{1}{2} \delta_{ij} B^2). Participants clarify that the integral of the stress tensor over the infinite hemisphere vanishes, allowing the focus to remain on the plane equidistant from the charges to determine the force. The integral over the plane is critical for calculating the force on one charge due to the other.

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stunner5000pt
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Griffiths problem 8.4
1. Homework Statement

Consider 2 eual point charges q, separated by a distance 2a. Construct the plance equidistnace form the two charges. By integrating Maxwell's stress tensor over the plane determine the force of one charge on the other.


Homework Equations


[tex]T_{ij} = \epsilon_{0} (E_{i}E_{j} - \frac{1}{2} \delta_{ij} E^2) + \frac{1}{\mu_{0}} (B_{i}B_{j} - \frac{1}{2} \delta_{IJ} B^2)[/tex]
where i and j are coordinates

The Attempt at a Solution


This does bear some similarities to the case for a uniformly charged solid hemisphere.. except in this case the bowl part is at z = infinity

suppose we had a poin charge located at z= a and another at z=-a and the plane is the XY plane.
the hemisphere shouldn't contribute to anything E=0 at z=infinity... right??

i am just wondering... how does this stress tensor 'work' for this problem

do we calculate the stress tensor on the plane (infinite plane) but how would that tell us anything about the electric field at a point beyond this plane??

thanks for your help
 
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stunner5000pt said:
Griffiths problem 8.4
1. Homework Statement

Consider 2 eual point charges q, separated by a distance 2a. Construct the plance equidistnace form the two charges. By integrating Maxwell's stress tensor over the plane determine the force of one charge on the other.


Homework Equations


[tex]T_{ij} = \epsilon_{0} (E_{i}E_{j} - \frac{1}{2} \delta_{ij} E^2) + \frac{1}{\mu_{0}} (B_{i}B_{j} - \frac{1}{2} \delta_{IJ} B^2)[/tex]
where i and j are coordinates

The Attempt at a Solution


This does bear some similarities to the case for a uniformly charged solid hemisphere.. except in this case the bowl part is at z = infinity

suppose we had a poin charge located at z= a and another at z=-a and the plane is the XY plane.
the hemisphere shouldn't contribute to anything E=0 at z=infinity... right??

i am just wondering... how does this stress tensor 'work' for this problem

do we calculate the stress tensor on the plane (infinite plane) but how would that tell us anything about the electric field at a point beyond this plane??

thanks for your help
1. You can show that the integral of T over the hemisphere vanishes.
The area of the hemisphere ~R^2, wjhile E^2~1/R^4.
2. Just do the integral over the plane: [tex]\int{\bf\hat k\cdot T}[/tex].
 
The total force of a collection of charges enclosed in a volume [tex]V[/tex] is given by

[tex]\mathbf{F} = \frac{d\mathbf{P}}{dt} = \oint_V T_{ij} dS_j[/tex].

This means that integrating over a closed volume containing one of the particles in your problem yields the force on that particle. The integral you are asked to calculate is the integral over the plane between the two particles, but you can allways enclose the integral by adding the integral over the infinite hemisphere (which is zero, as Meir Achuz pointed out above).

I hope this is clear.
 

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