0
$\begingroup$

I have been thinking about the idea given below

Consider the following situation

I have a rod which has a mass $m$ and hence does have the virtue to perform work against the gravitational field of the earth. I will for now neglect the air resistance and any such dissipative forces. I will now charge the rod via induction(i.e using another already charged rod) and then ground one of its ends simultaneously whilst in the process of charging . So I assume the rod has charge $q$. Now if I raise the rod some till height and let it go, then The rod will fall freely.

Now looking at the situation can we say current is flowing in the vertical direction and not along the rod? Also is the potential difference caused by gravity here? Can this be compared to electric potential.

$\endgroup$
4
  • $\begingroup$ Are you including consideration of the Earth's electric field and magnetic field? $\endgroup$
    – Farcher
    Commented Aug 2, 2023 at 8:34
  • $\begingroup$ @Farcher No, I wasn't $\endgroup$ Commented Aug 2, 2023 at 8:43
  • $\begingroup$ Why is this question closed $\endgroup$ Commented Aug 2, 2023 at 10:46
  • $\begingroup$ @Qmechanic What additional details do I need to give for this question? $\endgroup$ Commented Aug 2, 2023 at 11:04

2 Answers 2

1
$\begingroup$

Yes, there is a current flowing as the rod falls. Current is when charge moves. There is charge moving, so there is a current. There is nothing illusory about it. You would detect a magnetic field around the rod as it falls in accordance with Ampere's Law etc. However, note that static charges are typically very small (picocoulombs to nanocoulombs), so any current associated with a statically charged object being mechanically flung around is also very small. Thus all effects related to this current would be tiny and hard to measure. (The produced magnetic field mentioned before would be tiny. The deflection of the rod by an external magnetic field would be tiny. Etc.)

The overall potential energy of each charge in the rod is $U=qV_e+mV_g,$ where $V_e$ is the electric potential, which is constant inside the rod, and $V_g=gh$ is the gravitational potential, which does vary along the rod, in the direction that it's falling. The variation in $V_e$ is not providing the force for the charges to move, but the variation in $V_g$ is. So yes, gravity is producing the potential difference that makes the current appear. But remember that this potential should be considered as distinct from the electric potential.

$\endgroup$
0
$\begingroup$

I think your specific case has already been suggested by Lenz:

https://en.wikipedia.org/wiki/Lenz%27s_law

Note that the only feasible way to charge rod via induction and retaining its magnetic/electrostatic properties is by leaving an active field interacting with the rod. The moment that you remove the current or magnetic field from the rod it will discharge and it will be just another rod falling freely from the sky.

$\endgroup$
4
  • $\begingroup$ I was'nt considering magnetic fields. I was thinking if the change in position of charges constitutes current also? $\endgroup$ Commented Aug 2, 2023 at 7:19
  • $\begingroup$ Charging by induction can be done and kept without keeping a charged object nearby. You temporarily ground the object under the influence of the external field, this forces extra charge onto the object, removing the ground leaves it with surplus charge thereby charging the object even when the external field isn't present. $\endgroup$
    – Triatticus
    Commented Aug 2, 2023 at 7:21
  • $\begingroup$ @Triatticus sorry edited $\endgroup$ Commented Aug 2, 2023 at 7:22
  • 1
    $\begingroup$ That wasn't to you but to the answerer. $\endgroup$
    – Triatticus
    Commented Aug 2, 2023 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.