- #1
PhizKid
- 476
- 1
Homework Statement
Given a disc of radius 'R' with a charge 'h' above the center of the disc, find the electric field on the charge.
Homework Equations
##E = \frac{kQ}{d^2}##
The Attempt at a Solution
I can find the charge density from a ring then integrate over R:
##\sigma = \frac{Q}{A}##
##Area_{ring} = 2\pi r*dr##
##Q = \sigma * 2\pi r*dr##
Since the horizontal components cancel out, I'll take the cosine angle:
##E_y = \frac{kQ}{(h^2 + r^2)} * \frac{h}{\sqrt{h^2 + r^2}}##
##E_y = \frac{kh2\pi \sigma dr}{(h^2 + r^2)^{\frac{3}{2}}}##
##E_y = kh\pi \sigma \int_{0}^{R} \frac{2r dr}{(h^2 + r^2)^{\frac{3}{2}}}##
##E_y = kh\pi \sigma [\frac{-2}{\sqrt{h^2 + r^2}}]##
Integrate from 0 to R:
##E_y = kh\pi \sigma [\frac{-2}{\sqrt{h^2 + R^2}} + \frac{2}{h}]##
Re-substitute ##\sigma = \frac{Q}{A}##:
##E_y = kh\pi \frac{Q}{A} [\frac{-2}{\sqrt{h^2 + R^2}} + \frac{2}{h}]##
##E_y = kh\pi * \frac{Q}{2\pi r*dr} * [\frac{-2}{\sqrt{h^2 + R^2}} + \frac{2}{h}]##
I'm now stuck with this 'dr' in the denominator. Not sure what to do from here