# B Electric field due to a charged sheet

1. Apr 25, 2016

### Dexter Neutron

When a plane sheet is given a charge Q the electric field at any point around it is given as $E = \frac{\sigma}{\epsilon_0}$. Thus mathematically it can be concluded that the electric field is constant everywhere and is independent of distance of the point from the sheet. But what logical explanation can be given for this?
Since electric field due to a point charge varies inversely with $r^2$ and a plane charged sheet is nothing but collection of such charges then why the electric field is constant?It would even imply that if I charge a plane sheet on earth its effect could even be felt on mars?
How is that possible?

2. Apr 25, 2016

### Staff: Mentor

Symmetry. Roughly speaking the surface for spherical symmetry increases as $r^2$ so the field decreases as $r{-2}$. Similarly the surface for cylindrical symmetry increases as $r{1}$ so the field decreases as $r{-1}$. In the case of planar symmetry the surface area does not change so the field does not decrease.

3. Apr 25, 2016

### Dexter Neutron

Thanks for your answer but you are indirectly explaining the same mathematical aspect. I know that the effective area of the guassian surface around the sheet would always remain same as that of the sheet thus the electric field is independent of distance of the point from the sheet but what I want is LOGICAL explanation.
We can consider any point charge on the sheet which would exactly behave as any other point charge and thus the independent field of this charge would vary inversely with $r^2$ but what causes the field to regain its strength as distance increases so as to maintain the constancy of the electric field i.e. what causes the field to not decrease as distance increases?

4. Apr 25, 2016

### pixel

First of all, the equation you present (which may be missing a factor of 2 in the denominator) is for an infinite plane charged sheet and that is the key to understanding the result. Now "infinite" is always an approximation, so if we want the field strength in a small area close to the sheet in our lab, a sheet that is 5 ft x 5 ft might be a good approximation. As to your question about Mars - the sheet would have to be millions of miles on a side to be considered "infinite" in that case and for the result to still hold - since the equation involves the charge density, that means a lot more charge will be on the large plate and that's why the field strength will still be what it is at Mars.

From symmetry, the electric field at a point must point normal to the surface of the plane as there is always charge to one side and the other to cancel out any component parallel to the surface (as long as the sheet is infinite). As to why the field strength is constant - when you're close to the sheet, the contribution from most of the charge has a small component normal the sheet. As you go farther away, the contribution of those areas is now more in the normal direction but of course the areas are further away. It all works out to give a constant field strength, again provided the sheet is "infinite." See for example the following, where they do the calculation of adding the contributions from each part of the sheet: http://mlg.eng.cam.ac.uk/mchutchon/chargedPlanes.pdf