Question

Determine the force (in the lab frame) between two charges moving with the same velocity V. What is the ratio of transverse-to-longitudinal force? Explain what happens to this ratio when the two charges lie on a line parallel to V or perpendicular to V?

Answer 1

When two charges are moving with same velocity then the force experienced by them will be electromagnetic force between those two electric charges.

This problem was briefly considered by FitzGerald in 1882 , as a prelude to discussion of
the force on a moving electric charge due to a magnet that has the same velocity. FitzGerald
showed that the force is zero in this lab-case.These frame arguments were a precursor to a simpler, relativistic argument that in the rest frame of a charge and comoving magnetic there would be no magnetic force on the charge, and hence the lab-frame force is also zero.
Professor Rowland has shown experimentally that a quantity of electricity moving acts like an electric current. This had been assumed by many physicists. It follows that two quantities of electricity moving in the same
direction with the velocity of light, would have no action on one another, their electrostatic
action being balanced by an equal and opposite electrokinetic action. As it is very unlikely
that anything depends on absolute motion, the motion here spoken of must be with respect
to something, and this can hardly be any other thing than the ether in space.
FitzGerald did not support his claim, which we now examine from a more “relativistic”
perspective.
We consider a pair of electric charges q whose separation is d0 in their common rest frame, where the electric field of the first at the position of the second is (in Gaussian units)
E0 = q*d2/dˆ0.   (1)
and the electrical force on the second charge is,
F0 = qE0 = q2*d2/dˆ0. (2)

When the charges are in motion with velocity v perpendicular to the line of centers d0, the
electric field of the first at the position of the second is

E = γE0 (v ⊥ d0),

Rowland showed that a moving electric charge is a kind of magnet, and the (lab-frame)
magnetic force of a moving charge on a comoving electric charge tends to cancel the (lab-frame) electric force between them, when the velocity v is perpendicular to the line of centersof the charges.

What is the ratio of transverse to longitudinal force?

Ans:

Poisson ratio is the ratio of transverse contraction (or expansion) strain to longitudinal extension strain in the direction of stretching force. Tensile deformation is considered positive and compressive deformation is considered negative. The definition of Poisson ratio contains a minus sign so that normal materials have a positive ratio.

n= -e(trans)/e(long)

What happens when the two charges lie on a line parallel to v or perpendicular to v?

Ans:

We consider a pair of electric charges q whose separation is d0 in their common rest

frame, where the electric field of the first at the position of the second is (in Gaussian units)
E0 = q*d2/dˆ0,. (1)
and the electrical force on the second charge is,
F0 = qE0 = q2*d2/dˆ0. (2)

When v ⊥ d0( perpendicular)
When the charges are in motion with velocity v perpendicular to the line of centers d0, the
electric field of the first at the position of the second is
E = γE0 (v ⊥ d0), where γ = 1/sq root (1 − v2/c2) , (3)
and the lab-frame magnetic field at this point is3
B = v/c× E = γ*(v/c)× E0 (v ⊥ d0). (4)
The lab-frame Lorentz force on the second charge is then, noting that d · v = 0,
F = qE0/γ (v ⊥ d0). (5)
As v approaches c, γ → ∞, and F → 0.

As Rowland showed a moving electric charge is a kind of magnet, and the (lab-frame)
magnetic force of a moving charge on a comoving electric charge tends to cancel the (lab-frame) electric force between them, when the velocity v is perpendicular to the line of centers of the charges. In the limit that v → c, this cancelation is complete.

Now for v parallel to d0
However, if the velocity v is parallel to the line of centers of the charges, then the lab-frame
electric field of the first at the position of the second is,
E = E0 = q*d2/dˆ0 = γq/d2 dˆ (v parallel d0)
where the lab-frame separation of the charges is d = d0/γ due to the Lorentz contraction,
while the lab-frame magnetic field at the second charge is zero, according to the first form
of eq. (4). Then, the lab-frame force on the second charge is
F = γ2( q2/d2 d)(v parallel d0)
which diverges as v → c.

Thus, the lab-frame force goes to zero as v → c only if v is perpendicular to the line of centers
d0 of the charges, while FitzGerald’s claim seems to be that the force would be independent
of v · d0.