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## Homework Statement

A disk is located at 0<r<1, z = 1. It has a uniform charge distribution of 200pC/m^2. A 30 [tex]\mu[/tex]C point charge is located at the origin. Determine the force acting on the point charge due to the electric dield produced by the disk.

## Homework Equations

[tex]E = \frac{F}{Q}[/tex]

[tex]F = \frac{kQ_{1}Q_{2}}{R^{2}}[/tex]

[tex]dQ = \rho_{s}ds[/tex]

[tex]Q = \int \rho _{s}ds[/tex]

## The Attempt at a Solution

I know the idea behind finding the solution but I can't get the right answer. Thinking back to my calculus classes I need to take very very small peices to make the integral. One of the problems that I keep running into is that the book I'm using seems to keep everything in Cartesian and at last minute without telling you they switch to polar. I think that this problem should be easier if it was in polar so I am attempting that.

The first thing that I need to do is find the E-Field. If I take a very small peice of disk the area would be

da = r * dr * d(theta)

then I use this to find the charge

[tex]dq = \rho_{s} * d\rho * d\Theta[/tex]

Then I plug that into the e-field equation.

[tex]dE = \frac{\rho_{s}*\rho d\Theta d\rho}{4\pi\epsilon_{0}R^{2}}[/tex]

Now I need to find R wich is the distance from the point of observation from the charge. This is what I think is causing most of my problem. The book says to ignore all vectors other then the Z vector but, it never works out. Here is what I'm doing.

R is going to be the point of observation minus whereever the charge is. So

([tex]R= (0-\rho a_{\rho}-1a_{z}[/tex]

I need the magnitude of this

[tex]\sqrt{-\rho-1}^{2}[/tex]

Now plug that into the E-field equation along with multiplying by the unit vector and integrating to find the actual E-Field I get the following Equation

[tex]E = \int\int\frac{\rho_{s}*\rho d\Theta d\rho(-\rho a_{\rho}-1a_{z})}{4\pi\epsilon_{0}\sqrt{(-\rho^{2}-1^{2}})^{3/2}}[/tex]

This gives me basically two sets of integrals one for each seperate componet of the unit vector. This is what my final equation looked like before I plugged in the numbers. I already taken out the integral of theata since the equation didn't depend on it and so it equalled 2pi.

[tex]\frac{-\rho_{s}}{2\epsilon_{0}}(\int\frac{\rho^{2}}{\sqrt{\rho^{2}+1}^{3/2}}d\rho a_{\rho}+\int\frac{\rho}{\sqrt{\rho^{2}+1}^{3/2}}d\rho a_{z})[/tex]

I end up with .233887 for the first integral and .3784 for the second integral. Then multiply by the constants out front and my final answer is 2.701 in the A_p and 4.279 in the A_z. Wich is not the correct answer.

The Answer is supposed to be 99.24 in the Az