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## Homework Statement

A rod with length L has a charge density of

*Ax*where

*A*is a constant and

*x*is measured from the center of the rod, with positive to the right. Set up the integral to find the electric field at point P (which is at a distance

*h*above the center of the rod) but do not evaluate. Be sure to tell the direction of the net electric field. Also find the total charge on the rod.

## Homework Equations

dEp = kdq/r

^{2}where k = 8.99E9

## The Attempt at a Solution

So for the integral, I set dq = Ax*dx and r = [tex]\sqrt{L^2/4 + h^2}[/tex] I stated that, due to symmetry, the net electric field on the x-axis should be equal to 0, though I don't know if I can make that assumption since I don't know the charge on the rod? Regardless, I ended up getting this integral:

Ep = kA/(L[tex]^2[/tex]/4 + h[tex]^2[/tex]) [tex]\int { xdx}[/tex] in the +y direction. The integral is from 0 to h, as well. Not sure if those are the correct bounds to be using, but it made sense to me.

From there I figured the only way I could calculate the charge is by solving the integral and then solving for Q:

Ep = kAx[tex]^2[/tex]/(L[tex]^2[/tex]/2 + h[tex]^2[/tex]/2) from 0 to h

Ep = 2kAh[tex]^2[/tex]/(L[tex]^2[/tex] + h[tex]^2[/tex]) and since [tex]\lambda[/tex] = Ax and the units for [tex]\lambda[/tex] = N/C you can figure out that the units for A must be C/m[tex]^2[/tex] so A = Q/L[tex]^2[/tex]

Ep = 2kQh[tex]^2[/tex]/(L[tex]^4[/tex] + h[tex]^2[/tex]*L[tex]^2[/tex])

L[tex]^4[/tex] + h[tex]^2[/tex]*L[tex]^2[/tex] = 2kQh[tex]^2[/tex]

Q = L[tex]^4[/tex] + h^2[tex]^2[/tex]*L[tex]^2[/tex]/2kh[tex]^2[/tex]

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So I am almost 100% sure that the way I solved for charge will make any physicist cry because, honestly, it just

*looks*wrong. Besides, I believe that the way I solved this here also assumes that the Electric Field at P is equal to 1. If anyone can help me through this problem and help me understand my mistakes I would greatly appreciate it. It's weird that this problem is giving me trouble when other things like charged arcs are easy for me to grasp.

Thanks in advance!

-Kevin