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I Potential due to a uniformly charged flat disk

  1. May 6, 2017 #1
    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. jcsd
  3. May 6, 2017 #2

    Dr Transport

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    check your integral......
     
  4. May 7, 2017 #3
    ##-\displaystyle\int^{(0,y, 0)}_{(0,0,0)} {E \cdot dx}##
    Are you saying about -ve sign ?
     
  5. May 7, 2017 #4

    vanhees71

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    No! Check the integral for the potential. There's something missing in your transformation to polar coordinates!
     
  6. May 7, 2017 #5
    Oh right, I guess then

    ## \int E \cdot \sqrt{dr^2 + r^2 d\theta^2}## would be correct ?
     
  7. May 8, 2017 #6

    vanhees71

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    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
    $$\rho(\vec{x})=\sigma(x,z) \delta(y) \quad \text{with} \quad \sigma(x,z)=\begin{cases} \frac{Q}{\pi a^2} & \text{for} \quad x^2+z^2<a^2, \\
    0 & \text{for} \quad x^2+z^2>a^2.\end{cases}$$
     
  8. May 8, 2017 #7
    I don't know Green's function. :(. Is there any way around it ?
     
  9. May 8, 2017 #8

    vanhees71

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    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,
    $$\delta^{(3)}(\vec{x}-\vec{x}')=\delta(x-x')\delta(y-y') \delta(z-z').$$
     
  10. May 8, 2017 #9
    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
  11. May 9, 2017 #10

    vanhees71

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    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.
     
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