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simulation of a test charge experiencing force in a vacuum.

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simulation of a test charge experiencing a force when a neutral body that was introduced faces a little polarisation.

I had just posted a similar question on physics stackexchange, however, I didn't understand that when we are placing the neutral body are we seeing how much electric force is reduced from the original body(those group of 5 electrons) or overall force the test charge experiences. My school textbook says the electrostatic force is maximum in empty space but if we place matter it seems to be increasing.

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  • $\begingroup$ Where is this simulation coming from? $\endgroup$ – probably_someone Jun 30 at 19:22
  • $\begingroup$ @probably_someone phet.colorado.edu/en/simulation/charges-and-fields $\endgroup$ – Pranav K Jun 30 at 19:26
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    $\begingroup$ A single small neutral body somewhere between the source charge and test charge isn't the same as the set-up usually considered in textbooks: that is source charge and test charge both embedded in a medium filling the space between them, around them and beyond. $\endgroup$ – Philip Wood Jun 30 at 19:32
  • $\begingroup$ Did you look at the link I sent in my previous answer? $\endgroup$ – Bob D Jun 30 at 19:35
  • $\begingroup$ @Bob D, yes I saw the link, according to it the new electric field generated by the polarisation should reduce net effective electric field but that means the test charge should have been experiencing less electric force but in the simulation it was increasing and that's why posted this question. $\endgroup$ – Pranav K Jun 30 at 19:47
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I believe @Phillip Wood is correct in his comment about the filling of the space with the dielectric, and not just partially filling it. It is when the space is filled that the polarization of the dielectric reduces the net electric field within the filled space. In other words, in your example the reduction is only within your little dipole.

Regarding the conclusions you have reached about your simulation, perhaps you have been following the comments between @Phillip Wood and I. Your simulation shows that the potential at the location of the test charge appears to increase because of the dipole. This is true but the degree to which it increases depends on the ratio of the distance between the test charge and the dipole to the separation distance between the charges of the dipole. The greater that ratio, the more the dipole looks like a net charge of zero, and the less the increase in potential.

So unless you have already done so, you need to factor into your simulation the ratio of the distance between the test charge and the dipole to the distance between the charges in the dipole. The greater that ratio is, the less the influence of the dipole on the test charge.

In closing, I want to thank Philip Wood for his insight.

Hope this helps.

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  • $\begingroup$ If space is full of the dielectric the negative ends of dipoles are next to positive ends of adjoining dipoles, so the distant fields cancel, but for a small isolated body in the space between source charge and test charge, this cancellation doesn't take place. I think that's the gist of it. $\endgroup$ – Philip Wood Jun 30 at 20:30
  • $\begingroup$ @PhilipWood could you elaborate on "but for a small isolated body in the space between source charge and test charge, this cancellation doesn't take place." Do you mean the yellow test charge in the OP diagram actually does see the dipole as having a net positive charge? $\endgroup$ – Bob D Jun 30 at 21:05
  • $\begingroup$ @PhilipWood In other words the dipole causes an increase in potential at the location of the test charge? $\endgroup$ – Bob D Jun 30 at 21:07
  • $\begingroup$ "Do you mean the yellow test charge in the OP diagram actually does see the dipole as having a net positive charge?" No, I meant that the dipole's negative and positive ends aren't next to positive and negative ends respectively of neighbouring dipoles stretching in a chain between source charge and test charge, as they would be in a continuous medium. For the isolated dipole in the lower diagram the positive end is nearer than the negative end to the test charge, so you'd expect the source's field at the test charge to be augmented. Not so for the continuous medium, owing to cancellation $\endgroup$ – Philip Wood Jun 30 at 21:25
  • $\begingroup$ of fields due to opposite charges in neighbouring dipoles. $\endgroup$ – Philip Wood Jun 30 at 21:26

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