# Electric field on a charge some height above the center of a disc

PhizKid

## 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

Homework Helper
Gold Member
Is the disc charged? Is the charge above the disc very small?

Staff Emeritus
Homework Helper

## 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}##
Think about what Q and A represent here.

##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}}##
You can't have lone differentials sitting around. You should have written
\begin{align*}
dA &= 2\pi r\,dr \\
dQ &= \sigma\,dA = 2\pi \sigma r \, dr \\
dE_y &= \frac{k\,dQ}{h^2+r^2} \frac{h}{\sqrt{h^2+r^2}} = \frac{2\pi hk \sigma r \, dr}{(h^2+r^2)^{3/2}}
\end{align*} If you have an infinitesimal quantity on one side of the equation, you have to have one on the other side as well.

##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}}}##
Look at those two lines. According to what you wrote,
$$\frac{kh2\pi \sigma dr}{(h^2 + r^2)^{\frac{3}{2}}} = kh\pi \sigma \int_{0}^{R} \frac{2r dr}{(h^2 + r^2)^{\frac{3}{2}}}$$ because both sides of the equation equal ##E_y##. The integration apparently does absolutely nothing.

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
You need to think about what Q and A represent. Q is the charge of what? A is the area of what?

1 person