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Gauss's law and Moving Charges

  1. Jun 19, 2012 #1
    I am confused as to how Gauss's law can hold for moving charges. When discussing how to use Gauss's law to calculate the charge of a moving object, my textbook says:

    "At the instant the moving charge passes the center of the sphere, the force on each test charge is measured, and the average of the force magnitudes is used to compute Q."

    So we are using a surface in the "stationary" or "lab" frame, to calculate the flux integral.

    By taking for granted that the surface integral is the same whether the object is moving or not, and then using conservation of charge, the book then derives the relativistic transformations of the electric field.

    But If the perpendicular component of the electric field of a moving charge at any point is greater, and the parallel component is unchanged, shouldn't E be greater almost everywhere? That would seem to make the flux through a given surface greater for a moving charge.
     
  2. jcsd
  3. Jun 19, 2012 #2
    The flux through a given plane would indeed change. That's why you always use a closed surface. The components at various points along the surface will change, but the surface integral is invariant. It also makes sense: it's still the field of one charge right?
     
  4. Jun 20, 2012 #3
    Okay, I think I figured it out. Let me know if this makes any sense.

    Let's say you measure the electric field a certain distance x from a stationary charge. Then, you perform the same measurement on an identical, but moving, charge. This measurement is made at the same distance from the charge (in your frame) and is made on a line parallel to the axis of motion of the charge.

    The electric field you measure for the moving charge will actually be weaker than the field you measured for the stationary charge. I know that the transformation law says that E' = E parallel to the motion, but this only applies if you make the distance adjustments for the different frames of reference. Now if I had adjusted the distance from which I made the second measurement so that the coordinates corresponded with the first measurement's coordinates (taking the particle as the origin, you would measure closer to the particle) by a lorentz transformation, THEN E' = E applies.

    Basically you'd have to measure at x' instead of x in order to measure equal fields.

    Which means that on some parts of a Gaussian surface the flux is weaker than for the stationary charge, which compensates for the stronger transverse field, and Gauss's law still holds.

    Is this correct?
     
  5. Jun 20, 2012 #4
    Yep, it seems correct. If I understood correctly you assume that the observer is at rest right?

    This I didn't really understand. What do you mean by "the coordinates corresponding with the first measurements coordinates"?

    If you mean that, if the observer is on the same frame as the charge, then he measures the same as when the charge was not moving in respect to his previous frame, it's correct.

    An applet you may find useful:

    http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html
     
  6. Jun 20, 2012 #5
    Yeah, I said that awkwardly, didn't I? Basically, before I posted this thread I was under the impression that measuring the electric field at a given distance from a moving charge and a stationary charge, both times in your frame, (provided you measure parallel to the movement) you would get the same result. I now think that this isn't true. You would actually measure a weaker field for the moving charge. What E' = E tells us is that if you measure in the particle frame, then transform your coordinates into the "stationary" frame and perform the measurement again, you will get the same result. But you actually now appear to be closer to the charge.

    Sound right?
     
  7. Jun 21, 2012 #6
    Pretty much. It simply means that if you stay still at point A on both cases, you will measure a different value when the charge passes from point B, than if you were measuring a still charge at point B. This depends on the relative velocity between you and the charge, therefore the actual intensity you measure depends on the coordinate system. The transformation will simply transform your measurement for different coordinate systems.
     
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