# Electric field between 2 plates

1. Nov 11, 2009

### staetualex

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

2. Nov 12, 2009

### blitz.km

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.

Cheers

3. Nov 12, 2009

### 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?

4. Nov 12, 2009

### blitz.km

5. Nov 12, 2009

### ZapperZ

Staff Emeritus
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.

6. Nov 12, 2009

### staetualex

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.