- #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