Hello,

I am interested if ANYONE can point a mistake in my reasoning... I am constantly getting a result that is 2X greater than the textbook derivation.

The problem is this:
We have 2 CONDUCTING, infinitely large, oppositely charged plates (far away), each with a charge distribution of s1. Being a conductor, the charges are concentrated on the surface. So, in essence, EACH SIDE of the conductor has a charge distribution of s1.

Now, we take those two plates and bring them together VERY close, but keep them ||.

What happens (textbook explanation), is that due to attraction between the plates, the charges on the OUTSIDE of the plates migrate towards the inside, creating a NEW CHARGE DISTRIBUTION of s = 2xs1.

Fine, great... I get all of it.

But here is the catch:

We wish to calculate the electric field BETWEEN the plates.
MY ANSWER IS THIS:
The electric field due to ONE plate is E1 = s/epsilon0.
The electric field due to the OTHER is the same: E2 = s/epsilon0.
Since electric field is a VECTOR, the NET electric field is: E = E1 + E2 = 2 X s/epsilon0.

THE BOOK says this:
"With twice as much charge now on each inner face, the new surface charge density (s) on each inner face is twice s1. Thus, the electric field at any point between the plates has the magnitude:
E = 2s1/epsilon0 = s/epsilon0 (because s = 2s1)"

As you can see, my answer is TWICE the book's answer... But I think that the book is wrong... what IT does is calculate the electric field due to ONE plate (with the new surface charge density), but it does not add it... the formula for a CONDUCTING plate is E = s/e... NOT E = s/2e, which is for a NON-conducting...


PLEASE help me... I have spent HOURS on the internet... but the sites I have found do not clearly distinguish between PLATES and CONDUCTING PLATES.

Thanks,
Alek

The trouble is with the redistribution of charge on to one surface of each conductor. First, consider a nonconducting plane of charge. The E field comes out equally from both sides, and the strength is s/(2epsilon0). However, it is well known that the E field inside a conductor must be zero (under static conditions), so since the surface charge on a conductor sends equal E fields outside and inside the conductor, the field on the inside must be canceled out somehow. This can only mean that the E field from the charge density on the other side of the conductor (which also sends E both inwards and outwards) must be arriving from the other side with exactly the strength and direction to cancel out the field produced by the charge density on this side of the conductor. This ensures that the E field inside the conductor is zero, and the reasoning here actually is valid for ANY shape of conductor, not just a flat plate.

Now what about the field just outside the surface? In this case, the E field from the surface charge on this side of the conductor is ADDING with the E field from the surface charge on the other side of the conductor. This precisely doubles the field you see coming out of the surface. The reason is that the field coming out from this side's charge is +E and the field going into the conductor from this side's charge is -E (by symmetry and Gauss's law). Therefore, the field coming through from the charge on the other surface must have been exactly +E, so just outside the surface the E field is precisely double of what the charge on this surface produces. The conclusion is that the field you see coming out of the surface charge on a conductor is actually the contribution from BOTH surfaces.

Now put your two conducting planes near each other. As you said, the charges migrate to the near sides of the conductors. This is correct, and the surface charges are now s = 2s1. However, there is now NO CHARGE ON THE OTHER SURFACE of the conductors, so all you have from each surface is its own E field, which is 2s1/(2epsilon0) = s1/epsilon0. Summing the contributions from both plates gives 2s1/epsilon0 just as it should. Where you were tricked is that in your course, too much emphasis was placed on the surface of the conductor that you were looking at, and the contribution of the far surface was probably never mentioned.

Hope this helps!

--Stuart Anderson

can you give relevant diagram??

you said that E field is 2s1/(2epsilon0), here do you consider the side on which the charges reside as a sheet of charge??