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I Force on a stationary test charge from a moving charge

  1. Nov 30, 2018 #1
    Given a test charge, let's say on the vertical y axis. Another charge vertically below it moves past at a significant speed compared to c. From the Lorentz transformations (in most E&M books) the electric field and thus the force on the test charge will be greater than would be the static coulomb force it the moving charge was instead at rest below the test charge. Does this treatment give the total force?

    I ask because the moving charge is also said to have a magnetic field. A non moving test charge would of-course have no magnetic force on it if it is in a steady magnetic field. But the lower charge is moving by, so wouldn't its magnetic field be moving past the test charge and that moving magnetic field generate an electric field that would affect the test charge? Is this correct? If so, would this be a force in addition to the force mentioned above or does the above E force incorporate it?

    By-the-way, I took E&M decades ago with the text from Reitz and Milford (yuk). Well I did, Rusty I am.
  2. jcsd
  3. Dec 1, 2018 #2


    Staff: Mentor

  4. Dec 3, 2018 #3
    Well that parallels Feynman's treatment and I see how that works. My question actually revolved around the equation for the electric field from a moving charge. The text I was using is Purcell/Moran and I am sure Feynman has the same.

    The electric field of the moving charge was given as: E' = ( Q /4 pi eo r;^2) x (1-b^2) / (1-b^2sin^2 theta")^3/2. A relativistic equation. I wasn't sure if this would lead to the total force on the non-moving test charge. This field in the vertical direction is of course stronger than the field produced by a non moving charge. But regardless of its strength I questioned if this field as it moves past the non moving test charge would produce yet another field component - the moving charge having a magnetic field that in motion creates another E field. I suspect all is included in the above but I am not positive.
  5. Dec 4, 2018 #4

    Let's say there is a magnetometer next to the test charge. And let's say once a minute another charge passes the test charge.

    Okay now we have a pulsing current. When the current is increasing the reading on the magnetometer is going up, and an EMF is induced on the test charge.

    Let's say the passing charge is a charged metal ball , length contracted to half of its normal length. Well, that length contraction would double the rate of the change of the current, which would double the induced EMF, right?

    The aforementioned magnetic field, in the frame of the magnetometer, is what the magnetometer says it to be, the magnetometer does not know if it's a moving magnetic field or a still standing one. (We don't know or care either, that's my point.)

    The magnetic field momentarily stops changing when it's at its maximum, so induced EMF is zero at that moment.
  6. Dec 4, 2018 #5
    Well, that of course makes sense. But my question relates to the relativistic equation given for the increased electric field of the moving charge. That equation generates a graph of increased electric field in the vertical direction. So being a relativistic equation with the increased field due to motion does this equation give the total field and thus total force on the test charge? I suspect is does but I am not certain. The moving charge produces a magnetic field that will change in strength at the test charge location and a changing magnetic field will generate an electric field. Is that included in the relativistic equation given above?
  7. Dec 6, 2018 #6

    Hey, I recognize that formula in post #3, it's "Biot-Savart law for point charge", derived by Oliver Heaviside 1888.

    So let's see what Mr Heaviside says:

    He seems to be saying tat Faraday's law of induction is obeyed. (I don't really understand very much about that paper)

    Let's check one thing:

    E' = ( Q /4 pi eo r;^2) x (1-b^2) / (1-b^2sin^2 theta)^3/2

    When the passing charge is closest to the test charge the sin of theta is 1. If the v is 0.866c, then
    (1-b^2) / (1-b^2sin^2 theta)^3/2 = 2.0, so electric field is doubled.

    I agree with that result, electric field strength should double.
    Last edited: Dec 6, 2018
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