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

In summary, the answer to this problem is that the force on a charge located at z=a is given by the sum of the forces on the charges located at z=-a and z=infinity.
  • #1
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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  • #2
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].
 
  • #3
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.
 

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

What is force on a point charge?

Force on a point charge refers to the strength and direction of the force exerted on a charged particle by an external electric or magnetic field. This force is a vector quantity and is influenced by the magnitude and direction of the charge, as well as the strength and direction of the field.

How is the force on a point charge calculated?

The force on a point charge can be calculated using Coulomb's Law, which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for this calculation is F = (k * q1 * q2) / r^2, where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

What factors affect the force on a point charge?

The force on a point charge is affected by the magnitude and direction of the charge, the strength and direction of the external electric or magnetic field, and the distance between the charges. Additionally, the medium in which the charges are located can also impact the force.

How does the direction of the force on a point charge change?

The direction of the force on a point charge depends on the direction of the external electric or magnetic field. If the field is uniform, the force will be in the same direction as the field. However, if the field is not uniform, the force will vary in direction depending on the location of the charge within the field.

What is the difference between electric and magnetic forces on a point charge?

The main difference between electric and magnetic forces on a point charge is the type of field that causes the force. Electric forces are caused by electric fields, which are created by stationary charges, while magnetic forces are caused by magnetic fields, which are created by moving charges. Additionally, electric forces are always attractive or repulsive, while magnetic forces can only be attractive.

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