1
$\begingroup$

If a neutral metallic particle were to be dropped down, exactly in between two oppositely charged parallel plates, what would happen to it?

Motion-wise, does it deflect? Will it become charged?

Does the distance to the plate matter (e.j. its dropped closer to the positive plate)?

Thank you in advance!

$\endgroup$
5
  • $\begingroup$ $\vec{F} = q \vec{E}$. If $q = 0$ then $\vec{F} = 0$, so a neutral particle will feel no force no matter what. $\endgroup$ Apr 13, 2021 at 3:03
  • $\begingroup$ ohhh. what about induction? Wouldn't coming in near a charged object make the particle have charged sides? $\endgroup$ Apr 13, 2021 at 3:14
  • $\begingroup$ @moonlight-hollow Which cancels itself out because one side would be positive and other negative. $\endgroup$
    – DKNguyen
    Apr 13, 2021 at 3:18
  • $\begingroup$ It's true that if the particle is made up smaller positive and negatively charged constituent parts, the external magnetic field can polarize it and more complicated things can happen. However, in between two parallel plates, the electric field is constant throughout space, and in this case the overall charge will still just be 0. If one "part" has charge $+q$ and the other "part" has charge $-q$, then $\vec{F} = + q \vec{E} - q \vec{E} = 0$. If the electric field was varying throughout space, though, then the net force could be non zero. $\endgroup$ Apr 13, 2021 at 3:27
  • $\begingroup$ When the small conductor is very closed to one of the two plates. It will draw more intense field lines on the closed end of the conductor, and cause the local electric field deviate from the uniform field? $\endgroup$
    – ytlu
    Apr 13, 2021 at 4:17

4 Answers 4

1
$\begingroup$

If the electric field tension is high enough, it may ionize the neutral body and make a current, which is electrical charges motion along the electric field lines. Lightening is an example.

$\endgroup$
0
$\begingroup$

I'll try to get into a trivial case; The detail calculation is non-trivial.

Supposing you have a metal sphere and you put in the uniform field between two metal sheets. Then there will be an induced charge in a metal sphere which would be positive on one side and negative on the other side. Still, the sphere will not feel any force (?) as the field is considered to be uniform.

If you drop the sphere; in between two large plates, it just drops down without deflecting. If the field is non-uniform then there will be a deflection. So the distance between the plate and size of the plate does matter.

$\endgroup$
0
$\begingroup$

I think it depends on the fact that the particle is actually thought as an extended body and on the fact that the field is locally uniform for the particle or not (which can change the way in which the particle is polarized). Experimentally I can only think about the Millikan experiment, which employed small oil particles (not metal) and an electric field in which they were sprayed. They were deflected because, being originally neutral, the friction with a nozzle charged them. Without this workaround they wouldn't have been deflected (of course I am neglecting ionization through friction with air).

$\endgroup$
0
$\begingroup$

At rest the sphere is polarized and also the plates will be locally polarized. There is no net force on the sphere, but there is is a local force on the plates. Both the plates and the sphere will deform. When the sphere moves, the polarisation and mechanical deformation of the plates will travel with it. Inevitably losses are involved so that the fall of the sphere experiences friction. The amount depends on the material properties of both the plates and the sphere.

The polarisation may also break the symmetry of the system. Another effect to consider is the Casimir-Polder force, which may also break the symmetry.

The heat generated in the plates during the fall will cause a downward radiation pressure.

I may still have overlooked a few effects. Also I have assumed a vacuum.

$\endgroup$

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.