# I Potential due to a uniformly charged flat disk

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1. May 6, 2017

### Buffu

Let some point P on the y axis and let the disk of radius "a" to lie on the z-x plane perpendicular to y axis.
All charge elements in a thin ring shaped segement of the disk lie at same distance from P. If s denotes the radius of such a annular segment and ds is its width, then the area is $2\pi s ds$.

therefore $\displaystyle \phi(0,y,0 ) = \int {dq \over r} = \int^a_0 {2\pi \sigma ds \over \sqrt{y^2 + s^2}} = 2 \pi \sigma(\sqrt{y^2 + a^2} - y)$ Provided, y > 0 ,

I don't get how did we got $\int {dq \over r}$ should not it be $\displaystyle\int^{(0,y, 0)}_{(0,0,0)} {E \cdot dx}$ ? where dx is the displacement vector.

2. May 6, 2017

3. May 7, 2017

### Buffu

$-\displaystyle\int^{(0,y, 0)}_{(0,0,0)} {E \cdot dx}$
Are you saying about -ve sign ?

4. May 7, 2017

### vanhees71

No! Check the integral for the potential. There's something missing in your transformation to polar coordinates!

5. May 7, 2017

### Buffu

Oh right, I guess then

$\int E \cdot \sqrt{dr^2 + r^2 d\theta^2}$ would be correct ?

6. May 8, 2017

### vanhees71

I don't understand what you want to do with $E$ here. Either you evaluate the potential, using the Green's function for the negative Laplacian, i.e., in Heaviside-Lorentz units
$$\phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|},$$
or you evaluate directly the field
$$\vec{E}(\vec{x})=-\nabla \phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x'} \frac{\vec{x}-\vec{x}'}{4 \pi |\vec{x}-\vec{x}'|^3}.$$
In your case of a disc you have
$$\rho(\vec{x})=\sigma(x,z) \delta(y) \quad \text{with} \quad \sigma(x,z)=\begin{cases} \frac{Q}{\pi a^2} & \text{for} \quad x^2+z^2<a^2, \\ 0 & \text{for} \quad x^2+z^2>a^2.\end{cases}$$

7. May 8, 2017

### Buffu

I don't know Green's function. :(. Is there any way around it ?

8. May 8, 2017

### vanhees71

There's a very simple physical argument to introduce the Green's function. You start from the fact that electrostatics is a linear theory:
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
The first equation lets you introduce a scalar potential,
$$\vec{E}=-\vec{\nabla} \phi,$$
and you are left with just one equation
$$-\Delta \phi=\rho.$$
You can easily solve it for a point particle sitting in the origin by just looking at the problem in spherical coordinates. By symmetry it's clear that $\phi=\phi(r)$, i.e., it depends only on $r$, because it's symmetric under rotations. This implies
$$\frac{1}{r} (r \phi)]''=0 \quad \text{for} \quad r \neq 0.$$
Integrating step by step gives the solution
$$(r \phi)'=A \; \Rightarrow \; r \phi=A r+B \; \Rightarrow \; \phi=A+\frac{B}{r}.$$
An additive constant is irrelevant for the physics, because what counts is anyway only
$$\vec{E}=-\vec{\nabla} \phi=\frac{B \vec{x}}{r^3}.$$
To get the constant $B$ we integrate $\vec{E}$ over a sphere of radius $a$ around the origin. This should give the total charge inside, which is $Q$:
$$Q=\int_{S_a} \mathrm{d}^2 \vec{f} \cdot \vec{E}=4 \pi B \; \Rightarrow \; B=\frac{Q}{4 \pi}.$$
So we have the solution for the potential and the field for a point charge in the origin:
$$\phi=\frac{Q}{4 \pi r}, \quad \vec{E}(\vec{x})=\frac{Q}{4 \pi r^3} \vec{x},$$
which, of course, is the Coulomb potential.

Now it's clear that for a charge at position $\vec{x}'$ you get
$$\phi=\frac{Q}{4 \pi |\vec{x}-\vec{x}'|},$$
because the equations are invariant under translations.

Since the equations are linear, the potential for several charges $Q_j$ sitting at position $\vec{x}_j'$ is given by superposition:
$$\phi(\vec{x})=\sum_{j} \frac{Q_j}{4 \pi |\vec{x}-\vec{x}_j'|}.$$
For a continuous charge density $\rho(\vec{x})$ you have "infinitesimal charges" at places $\vec{x}'$ of the amount $\mathrm{d} Q=\mathrm{d}^3 \vec{x}' \rho(\vec{x}')$, which you have to "sum", but the some becomes an integral when making the volume elements smaller and smaller, this leads to
$$\phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
So we have found a function
$$G(\vec{x},\vec{x}')=\frac{1}{4 \pi |\vec{x}-\vec{x}'|},$$
which defines an integral operator, inverting the negative Laplace operator in the Poisson equation:
$$-\Delta \phi(\vec{x})=\rho(\vec{x}) \; \Rightarrow \; \phi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d} ^3 \vec{x} G(\vec{x},\vec{x}') \rho(\vec{x}').$$
Such an operator is called a "Green's function" (named after George Green, who developed this approach to solve linear partial differential equations in the 19th century).

A modern definition for the Green's function of the (negative) Laplacian is
$$-\Delta_{\vec{x}} G(\vec{x},\vec{x}')=\delta^{(3)}(\vec{x}-\vec{x}'),$$
where $\delta^{(3)}$ is the 3D Dirac-$\delta$ distribution,
$$\delta^{(3)}(\vec{x}-\vec{x}')=\delta(x-x')\delta(y-y') \delta(z-z').$$

9. May 8, 2017

### Buffu

It was way easier to understand than it looked. You wrote it so nicely. Thanks for the effort.

Just I did not get two things,

First,
In sperical coordinates $\displaystyle -\nabla f = {\partial f \over \partial r}\hat r$ when $\theta$ and $\varphi$ is constant.

So, $\displaystyle-\vec{\nabla} \phi=-{ \partial (A + B/r) \over \partial r}\hat r = {B \over r^2}\hat r$

I guess to muliplied and divided by the magnitude to get $\vec r$ instead of $\hat r$, so you got $1/r^3$ in denominator. Correct ?

Second one is very stupid,

Why does integrating the field over the surface gives charge ?

sorry this one is really stupid but I can't understand it.

Last edited: May 8, 2017
10. May 9, 2017

### vanhees71

Yes, indeed. If you write
$$\hat{r}=\frac{\vec{x}}{r}$$
you get
$$-\vec{\nabla} \phi=\frac{B}{r^2} \hat{r}=\frac{B}{r^2} \frac{\vec{x}}{r}=\frac{B \vec{x}}{r^3}.$$
The second thing follows from Gauss's Law (one of the Maxwell equations),
$$\vec{\nabla} \cdot \vec{E}=\rho.$$
We assume that the only charge is the charge $Q$ concentrated on the origin (point charge as an idealized classical model of a little charged body). Now use Gauss's theorem to any volume $V$ with the origin in its interior and $\partial V$ its boundary surface. Then you get
$$Q=\int_V \mathrm{d}^3 \vec{x} \rho=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}.$$
This is Gauss's Law in integral form.