AbhiFromXtraZ said:
Thanks for the link. But one thing I'm not getting clearly is E=(sigma/epsilon) when fiele is in one side of the plate and E=(sigma/2epsilon) when field is in both side...for the parallel plates field is in one side of each...so why the resultant field is (sigma/epsilon) and not (2sigma/epsilon) ??
And the problems I posted are already worked out and derived correct result...the capacy of a parallel plate capacitor of area A is (epsilon*A/d).
You didn't do that math did you? :)
put two infinite plates equal charge distribution, opposite charge, parallel to x-y plane, with the +q plate at z=+d/2 and the -q plate at z=-d/2 and remember that the electric field is a vector.
Field due to +q plate is
$$\vec E_+ = \left\{\begin{align} -\frac{\sigma}{2\epsilon_0}\hat{k} &:\; z>d/2\\ 0 &:\; z=d/2\\\frac{\sigma}{2\epsilon_0}\hat{k} &:\; z<d/2\end{align}\right.$$
Field due to -q plate is
$$\vec E_- = \left\{\begin{align} \frac{\sigma}{2\epsilon_0}\hat{k} &:\; z>-d/2\\ 0 &:\; z=-d/2\\-\frac{\sigma}{2\epsilon_0}\hat{k} &:\; z<-d/2\end{align}\right.$$ (caution: don't take my word for it: check!)
Now you do the math: the total field is $$\vec E = \vec E_+ + \vec E_-$$
... you should be able to confirm dauto's result:
dauto said:
The field of each plate is E = σ/(2ε). Both plates together produce a field twice as big E = σ/ε
Because the electric field is a vector:
Philip Wood said:
...outside the capacitor the fields cancel!
... but that is only for infinite plates. Real life capacitors have some field outside and the above is just an approximation.
Taking the approximation too literally can lead to odd results.
i.e. http://www.lhup.edu/~dsimanek/museum/advanced.htm
(3rd one down: 2nd of the "two puzzles" - try it before looking at the answer).
The path through this sort of confusion is to take things one step at a time checking as you go.