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

## 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

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

vela
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?