Question

Can someone help me in deriving the equation for the electric field due to a charged disk?

Answer 1

Hi guys! Try to work on that by yourselves. I can help later or comment on your formulae.

Answer 2

equation of electic feild due to a charged disk.

Answer 3

okay..i did it like this - i just wanna know where i went wrong

Answer 4

then you have to see it in your work nah..

Answer 5

sigma( area charge density) = dq/dA
or sigma * dA = dq;
=sigma * (2 pie r dr)
dE = 1/4 pie (ep) * (sigma* 2 pie r dr)/(r^2 + z^2)^2
along x axis, E will be zero
along Y axis;
dEsin theta comes out to be 1/4pie (ep) * (sigma*2 pie r^2 dr)/(r^2 + z^2)^3/2;
(put the value of sin theta as r/ sqrt (r^2 + z^2) and check that;
from here, when i integrate i get 1/2 (ep) * sigma r^2 * 2 pie r/(r^2+z^2)^3/2;

Answer 6

ep = epsilon

Answer 7

@goutham1995 pls use eqn editor

Answer 8

@AravindG - how?

Answer 9

on bottoof reply u have equation draw attach file,
click equation then type in

Answer 10

*bottom

Answer 11

\[ \int_{0}^{a}\frac{kq\sigma2\pi r dr \cos\theta}{r^2 + x^2}\]
\( \cos\theta = \frac{x}{\sqrt{x^2+r^2}}\)

Answer 12

\[\sigma/ 2\epsilon (1 - z/\sqrt{z^2 + R^2) is the answer

Answer 13

what--sorry it didnt print out properly -

Answer 14

sigma/2 (ep)( 1 - z/sqrt(z^2 + R^2)) should be the answer

Answer 15

\[\sigma/2\epsilon (1 - z/\sqrt{z ^{2 }+R ^{2}}\]