PlanetPhysics/Electric Field of a Charged Disk

Electric Field of a Charged Disk

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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   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

 

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  

cartesian coordinates:

 

cylindrical coordinates:

 

\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

   

therefore

   

substituting these relationships into (1) gives us

 

As usual break up the integration into the   and   components

{\mathbf z} component:

 

Since   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

 

using u substitution

     

with the limits of integration becoming

   

trasnforming the integral to

 

integrating

 

evaluating the limits

 

integrating again simply gives

 

  component:

 

If you cannot simply see how the   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   component, is to realize that   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  . An important result from cylindrical coordinates is the relation between its unit vectros and those of cartesian coordinates.

     

Plugging in the   into our integral

 

  component:

To make our job easier, let us first integrate  

 

Note how   can be taken out of integral, so we get

 

Evaluating the limits, gives us the result we expected.

 

  component:

 

integrating

 

which once again yeilds a zero.

 

Since the x and y components are zero

 

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

 

and rearranging

 

Superposition

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