# electricfield in disk

68 Points
13 years ago

Dear  satish,

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.

Starting with the general formula for a surface charge

 (1)

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:

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

 (2)

As usual break up the integration into the and components

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

All the best.

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