Potential due to a uniformly charged flat disk

[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.

 

[Dr Transport]

check your integral...

 

[Buffu]

check your integral...

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

 

[vanhees71]

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

 

[Buffu]

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

Oh right, I guess then

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

 

[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
$$\rho(\vec{x})=\sigma(x,z) \delta(y) \quad \text{with} \quad \sigma(x,z)=\begin{cases} \frac{Q}{\pi a^2} & \text{for} \quad x^2+z^2<a^2, \\
0 & \text{for} \quad x^2+z^2>a^2.\end{cases}$$

 

[Buffu]

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
$$\rho(\vec{x})=\sigma(x,z) \delta(y) \quad \text{with} \quad \sigma(x,z)=\begin{cases} \frac{Q}{\pi a^2} & \text{for} \quad x^2+z^2<a^2, \\
0 & \text{for} \quad x^2+z^2>a^2.\end{cases}$$

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

 

[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 let's 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,
$$\delta^{(3)}(\vec{x}-\vec{x}')=\delta(x-x')\delta(y-y') \delta(z-z').$$

 

[Buffu]

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

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 BBB we integrate ⃗EE→\vec{E} over a sphere of radius aaa around the origin. This should give the total charge inside, which is QQQ:

$$Q=\int_{S_a} \mathrm{d}^2 \vec{f} \cdot \vec{E}=4 \pi B \; \Rightarrow \; B=\frac{Q}{4 \pi}.$$​

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.

 

[vanhees71]

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.

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.