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My question has to do with this graphic:

enter image description here

I am perfectly fine accepting the fact that the object will move parallel to $\vec E$ and perpendicular to the surface of the sheet. Any attraction from the upper parts of the sheet will be canceled out by the bottom or vice-versa. But, what if the test charge was raised up a bit, such that it was at the upper-half of the sheet when we begin to examine its motion?

Then, there "will be more things pulling it down" (for lack of a better word) than trying to pull it up (as there is more electric field lines below the particle than above it), so will it still move off perpendicular to the surface of the sheet or deflect downwards?

My guess is it will still move perpendicular, based off intuition and also due to the fact that, although there is less electric field lines above the test charge than below, due to the fact that the electrostatic force is an inverse square law, it'll still manage to cancel out.

Am I correct? And if not, why not?

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  • $\begingroup$ -1 Not clear what you are asking. Is the sheet infinite? If not, why are the field lines all perpendicular to the surface? If it is infinite, and the field lines are all perpendicular, why do you think the test charge will be "pulled down" parallel to the plane sheet? The test charge always starts to move in the direction of the local E field. $\endgroup$ – sammy gerbil Jan 30 '18 at 0:02
  • $\begingroup$ Possible duplicate of Electric field of an infinite, uniformly charged layer with thickness $a$ $\endgroup$ – sammy gerbil Jan 30 '18 at 0:18
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You are making a valid argument. In the case that the sheet is not infinite but has a finite area, the electric field lines, at a distance from the charge sheet, will not be perpendicular to the sheet but "curve upwards" in the upper half and "curve downwards" in the lower half. The electric field will not be normal to the sheet there. When the charge sheet has an infinite area, the electric field will always be perpendicular to the sheet independently of where you position your test charge along the sheet.

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  • $\begingroup$ Right, because every single electric force trying to pull the charge down will be met with an equal and opposite electric force pulling it up, since there are infinitely many sources to cause an electric force. So, given it's finite, there will be curving for the charge? $\endgroup$ – sangstar Jan 30 '18 at 16:58
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    $\begingroup$ @sangstar - Yes, there will be! As long as you are not at the exact center of the finite charge sheet. You are completely right. $\endgroup$ – freecharly Jan 30 '18 at 17:05
  • $\begingroup$ Oh okay, so my intuition was correct. Thank you for clearing that up! $\endgroup$ – sangstar Jan 30 '18 at 17:19

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