Want to know correct answer for electric field + potential question

[altegron]

Homework Statement

http://img352.imageshack.us/img352/9444/enm1qi8.png

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

$$V = {k}_{e}\frac{q}{r}$$

$$E = {k}_{e}\frac{q}{r^2}$$

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:

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

And simplify/solve for d to get:

$$d = \frac{\sqrt{3}a}{3}$$

(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:

$$E = 2({k}_{e}\frac{q}{(\sqrt{a^2+x^2})^2})$$

and to get the x component multiplied that by:

$${E}_{x} = \cos\tan^{-1}\frac{a}{x}$$

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

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

$${E}_{x net} = \frac{2{k}_{e}q(\cos\tan^{-1}\frac{a}{x})(x^2) - {k}_{e}q(a^2+x^2)}{(a^2+x^2)x^2}$$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.

 

[Shooting Star]

3a) d should have two values.

3b) What happened to the field due to the charge (-q)?

 

[altegron]

Thanks for the reply!

Ah yeah that makes sense for (a). It should be plus or minus then.

For (b), would the E from -q be $-{k}_{e}\frac{q}_{x^2}$?

 

[Shooting Star]

and to get the x component multiplied that by:

$${E}_{x} = \cos\tan^{-1}\frac{a}{x}$$

Write cos theta = x/sqrt(x^2+a^2).

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

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

That is correct now.

 

[sadakaa]

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

can't you just replace $$(\cos\tan^{-1}\frac{a}{x})$$ with $$\frac{x}{\sqrt{(x^2+a^2)}}$$...or is that what shootingstar was saying