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.

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