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## Homework Statement

http://img352.imageshack.us/img352/9444/enm1qi8.png [Broken]

Three point charges are arranged on the y-axis as shown in the picture.

The charges are:

+q at (0,a)

-q at (0,0)

+q at (0,-a)

Any other charge or material is infinitely far away.

(a) Determine the point(s) on the x-axis where the electric potential due to this system of charges is zero.

(b) Determine the x and y components of the electric field at a point P on the x-axis distance x from the origin.

## Homework Equations

[tex]V = {k}_{e}\frac{q}{r}[/tex]

[tex]E = {k}_{e}\frac{q}{r^2}[/tex]

## The Attempt at a Solution

(a) The problem says the point(s) where potential is zero is must be on the x-axis so I know the y-coordinate is 0. I called the distance d, so my coordinate would be (d,0).

I make this equation:

[tex]0 = 2({k}_{e}\frac{q}{\sqrt{a^2+d^2}}) - {k}_{e}\frac{q}{d}

[/tex]

And simplify/solve for d to get:

[tex]d = \frac{\sqrt{3}a}{3}[/tex]

(b) For this part I guess I just need to express the E vector in terms of x, q, ke, and a.

Again I ignore any y components since the +q charges cancel in that direction.

To get the magnitude of the field vector from the (+) charges I plugged in the distance of P (using a right triangle with sides x and a) from the charge for r and doubled:

[tex]E = 2({k}_{e}\frac{q}{(\sqrt{a^2+x^2})^2})[/tex]

and to get the x component multiplied that by:

[tex]{E}_{x} = \cos\tan^{-1}\frac{a}{x}[/tex]

Then I find the net to be (including E from the (-) charge):

[tex]{E}_{x net} = 2(\cos\tan^{-1}\frac{a}{x})({k}_{e}\frac{q}{a^2+x^2}) - {k}_{e}\frac{q}_{x^2}[/tex]

[tex]{E}_{x net} = \frac{2{k}_{e}q(\cos\tan^{-1}\frac{a}{x})(x^2) - {k}_{e}q(a^2+x^2)}{(a^2+x^2)x^2}[/tex]

I feel ok about (a) but I might be wrong. As for (b), I have no idea if I'm right or not but I did try it multiple ways to get an equivalent solution. It just looks too complicated though. Please let me know if I have done these right.

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