PlanetPhysics/Electric Field of a Charged Disk

Electric Field of a Charged Disk
The Electric Field of a charged disk can teach us important concepts that you will see over and over in physics: superposition, cylindrical coordinates and non-constant basis vectors. To get a glimpse of the power of superposition, we will solve this problem the hard way first and then see how superposition can be a powerfull tool.

Let us calculate the Electric Field at a point P above the center of a charged disk with radius of R and a uniform surface charge density of $$\sigma$$ as shown in below figure.

\begin{figure} \includegraphics[scale=.8]{EfieldDisk.eps} \vspace{20 pt} \end{figure}

Starting with the general formula for a surface charge

$$ {\mathbf E} = \frac{1}{4 \pi \epsilon_0} \int \frac{\sigma(r')({\mathbf r} - {\mathbf r'}) da'}{\left |{\mathbf r} - {\mathbf r'} \right |^3} $$

choose a coordinate system. A disk clearly lends itself to cylindrical coordinates. As a refresher, the next figure shows the infinitesimal displacement, where we have the infinitesmal area $$da'$$

cartesian coordinates:

$$da' = dx'dy'$$

cylindrical coordinates:

$$da' = s'ds'd\phi'$$

\begin{figure} \includegraphics[scale=.5]{InfDis.eps} \vspace{20 pt} \end{figure}

The vectors to the source and field points that are needed for the integration in cylindrical coordinates

$${\mathbf r} = z \hat{z}$$ $${\mathbf r'} = s' \hat{\phi}'$$

therefore

$${\mathbf r} - {\mathbf r'} = z \hat{z} - s' \hat{\phi}'$$ $$ \left |{\mathbf r} - {\mathbf r'} \right | = \sqrt{s'^2 + z^2} $$

substituting these relationships into (1) gives us

$$ {\mathbf E} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R \frac{s'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}}\left ( z \hat{z} - s' \hat{\phi}' \right ) $$

As usual break up the integration into the $$z$$ and $$\phi$$ components

{\mathbf z} component:

$${\mathbf E}_z = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R \frac{z \hat{z}s'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $$

Since $$\hat{z}$$ is always in the same direction and has the same magnitude (unit vector), it is constant and can be brought out of the integration. Integrating the ds them

$${\mathbf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \int_0^R \frac{s'ds'}{\left ( s'^2 + z^2 \right )^{3/2}} $$

using u substitution

$$u = s'^2 + z^2$$ $$du = 2s' ds$$ $$ds = \frac{du}{2s}$$

with the limits of integration becoming

$$u(s'=0) = z^2$$ $$u(s'=R) = R^2 + z^2$$

trasnforming the integral to

$${\mathbf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \int_{z^2}^{R^2 + z^2} \frac{u^{-3/2}du}{2} $$

integrating

$${\mathbf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \left. \right |_{z^2}^{R^2 + z^2} -u^{-1/2}du$$

evaluating the limits

$${\mathbf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} \left ( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right ) d\phi' $$

integrating again simply gives

$${\mathbf E}_z = \frac{\sigma z \hat{z}}{2 \epsilon_0} \left ( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right ) $$

$${\mathbf \phi}$$ component:

$${\mathbf E}_{\phi} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R -\frac{s'^2\hat{\phi}'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $$

If you cannot simply see how the $$\phi$$ component is zero through symmetry, then carry out the integration. The key thing to learn here, and why it is not good to just skip over the $$\phi$$ component, is to realize that $$\hat{\phi}$$ is not constant throughout the integration. Therefore, one cannot bring it out of the integration. What needs to be done is to substitute in for $$\hat{\phi}$$. An important result from cylindrical coordinates is the relation between its unit vectros and those of cartesian coordinates.

$$ \hat{s} = \cos \phi \hat{x} + \sin \phi \hat{y} $$ $$ \hat{\phi} = -\sin \phi \hat{x} + \cos \phi \hat{y} $$ $$ \hat{z} = \hat{z} $$

Plugging in the $$ \hat{\phi}'$$ into our integral

$${\mathbf E}_{\phi} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R -\frac{s'^2 \left ( -\sin \phi' \hat{x} + \cos \phi' \hat{y} \right )ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $$

$${\mathbf x}$$ component:

To make our job easier, let us first integrate $$d\phi'$$

$${\mathbf E}_{\phi}^x = \frac{\sigma \hat{x}}{4 \pi \epsilon_0} \int_0^R  \frac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \int_0^{2\pi} \sin \phi'  d\phi'$$

Note how $$\hat{x}$$ can be taken out of integral, so we get

$${\mathbf E}_{\phi}^x = \frac{\sigma \hat{x}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}}  \left. \right |_0^{2\pi} - \cos \phi' $$

Evaluating the limits, gives us the result we expected.

$${\mathbf E}_{\phi}^x = \frac{\sigma}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 \hat{x} ds'}{\left ( s'^2 + z^2 \right )^{3/2}}  \left ( -1 - (-1) \right ) = 0   $$

$${\mathbf y}$$ component:

$${\mathbf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0^R  \frac{s'^2  ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \int_0^{2\pi} -\cos \phi'  d\phi'$$

integrating

$${\mathbf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2  ds'}{\left ( s'^2 + z^2 \right )^{3/2}}  \left. \right |_0^{2\pi} -\sin \phi' $$

which once again yeilds a zero.

$${\mathbf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2  ds'}{\left ( s'^2 + z^2 \right )^{3/2}}  \left ( 0 - 0 \right ) = 0   $$

Since the x and y components are zero

$$ {\mathbf E}_{\phi} = 0$$

Therefore, for a charged disk at a point above the center, we have

$${\mathbf E} = \frac{\sigma z \hat{z}}{2 \epsilon_0} \left ( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right ) $$

and rearranging

$${\mathbf E} = \frac{\sigma }{2 \epsilon_0} \left ( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right ) \hat{z} $$

Superposition
Before we can apply superposition to this problem, we need to calculate the electric field of a charged ring. This entry is coming soon.