Picture a positively charged glass rod from its end so that if you were to draw it on paper, it would appear as a circle. Now, suppose a negatively charged particle moves in a circular arc, on the same plane, around the rod. On paper, this should look like a circle orbiting another circle.

I know that the work done by the rod's electric field is zero because the force that the field exerts on the particle will always be perpendicular to the particle's direction of motion. So, my questions are:

  1. What does this mean for the trajectory of the particle? Will the trajectory change even if the work done by the field is zero?

  2. What does it even mean for the work to be zero? I know that work is the transfer of mechanical energy between systems, so I suppose this means that there is no energy transfer. So then, what does that mean in terms of the motion of the particle around the rod?

For context, these questions are based off of a reading I just did on the first section of the "Electric Potential" chapter of my second-semester college physics textbook.

  • $\begingroup$ You're neglecting EM radiation emitted by the particle's motion. $\endgroup$ – Daniel Griscom Nov 30 '15 at 21:44
  • 1
    $\begingroup$ One could ask the same question about planets orbiting the sun. Why should there be any transfer of energy (in the simplest case barring EM emission or gravitational wave emission)? $\endgroup$ – Jon Custer Nov 30 '15 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy