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Imagine two equally charged capacitor plates parallel to the x-y axis, whose area is large enough compared to the distance between them that fringe effects can be ignored. The bottom plate (at z=0) is + charged, and the top is - charged. The vector field E is therefore directed upward ('upward' in this case being the positive 'z' direction) from the + plate to the - plate, and its magnitude should be equal to [surface charge density/epsilon].

If we examine the Lorentz force q[E + vxB] experienced by a test charge situated between the two plates, and moving with constant velocity parallel to the plates as seen from the frame of reference in which the plates are at rest, relativistic theory tells us we should get the same total Lorentz force (albiet with different E and B values) when we examine it in the frame of the moving charge, in which the test charge is still, and the plates are in motion.

Well then, first let's look at the frame where the plates are at rest and the test charge moves. Because the plates are at rest, B must be zero, and therefore the vxB term must also be zero. The Lorentz force reduces to q[E].

But now let's look at the frame of the moving test charge. In this frame, there is a B-field due to the moving plates, but because we are moving with the test charge, it is at rest in this frame ... and therefore magnitude of B doesn't effect the force it experiences. The vxB term once again vanishes, leaving the Lorentz force to be q[E'], where E' means the electric field of the first reference [E] frame as transformed by the lorentz factor of [1/(1-v squared/c squared)].

But q[E] does NOT equal q[E']!

What is wrong with this analysis?

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# Relativistic transformation of the Lorentz force (E + v x B)

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