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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

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# Electric field due to || CONDUCTING plates

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