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Electric field due to  CONDUCTING plates 
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#1
Aug406, 10:11 PM

P: 3

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 NONconducting... 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 


#2
Aug506, 08:57 AM

P: 46

INTRODUCTION:Neither do i understand u, nor ur book.please try to b more specific next time.make headings and under each explain ur problem part by part. it becomes easier. wht is s? wht is s1? nothing is clear.But anyway as far as i can understand i am explaining.
CONCEPT BOOSTER:Since the charges on the two plates are opposite, let charge on one be +s and on the other be s. Now the plates are brought closer. now there won't be any charge on the outer surfaces of the plates. so, it would be like +s on the inner surface of 1st plate and s on the outer surface of the 2nd plate.the field due to a single surface of the 1st plate is s/(2epsilon) and not s/epsilon. if there is only one surface then it is always s/(2epsilon). so the field at a point in between is s/(2epsilon) due to inner surface of 1st plate and s/(2epsilon) due to inner surface of 2nd plate. so E1+E2 is s/epsilon. I think it matches your book's answer. CONCLUSION: Even i have learnt electric field due to  plates recently. I've tried my best to answer ur question. But if there is anything wrong in my answer, then please let me know.Most probably my answer is allright. 


#3
Aug506, 10:09 AM

Mentor
P: 41,477

Note that the field from a sheet of charge is given by E = s/(2epsilon0). You can also use this to calculate the field between the plates; in this case you must add the field from both sheets of charge: E(total) = 2*s/(2epsilon0) = s/epsilon0. Both methods give the same answer. (1) A single conducting plate with total charge density of s. The charge density on each surface is s/2. Using the formula for the field at the surface of a conductor gives us a field of (s/2)/epsilon0. Using the formula for the field from a sheet of charge gives us s/(2epsilon0) (we must use the total charge, not just the charge on one surface). The answers, of course, are the same. (2) Now we put two conducting plates near each other, each having the same total charge as before (but with opposite signs, of course). Now the charge density on each surface is doubled (as explained in your text), so using the formula for the field near a conducting surface gives you twice the field. And, using the other method, you also get twice the field since there are now two sheets of charge contributing to the field. 


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