Electric field between 2 plates

**staetualex** · Nov11-09, 10:54 AM

Why electric field between 2 charged plates is sigma(surface charge density)/(2epsilonO) permitivity of free space?

**blitz.km** · Nov12-09, 05:18 AM

You are wrong brother..
The electric field between 2 charged plated is sigma/epsilonO.. provided both the plates have equal and opposite charge..
This is proved using Gauss's Law.
If u haven't learnt about this concept.. search Wikipedia or any other good book.

Cheers

**staetualex** · Nov12-09, 05:21 AM

Yeah, sorry, one plate comes with sigma/2epsilon0, another comes with sigma/2epsilon0, add them together and we get sigma/epsilon0. Anyone wants to share why is the electric field sigma/2epsilon0?

**blitz.km** · Nov12-09, 05:26 AM

http://en.wikipedia.org/wiki/Gauss%27s_law

See this link.
It will help you learn about Gauss's Law..
which is the key to the derivation of electric field of a conductor.

**ZapperZ** · Nov12-09, 05:49 AM

Originally Posted by staetualex

Yeah, sorry, one plate comes with sigma/2epsilon0, another comes with sigma/2epsilon0, add them together and we get sigma/epsilon0. Anyone wants to share why is the electric field sigma/2epsilon0?

Your question is vague. Are you asking HOW one derives such an electric field for that configuration? If you are, then you already had your answer via Gauss's law. If you are asking on how to make such a derivation, then you need to make further elaboration on your academic background, i.e. have you been taught Gauss's law and how to apply it? And this needs to be done in the HW/Coursework forum.

Zz.

**staetualex** · Nov12-09, 05:54 AM

Finished high school, no college (19). I was studying Coloumb's law, electric flux and so on, and i stumbled upon that formula. Just wanted to know why the formula valid. That link, with gauss law, it may help me prove myself, just need to crunch the numbers.