Locus of all points due to a charge

1. Nov 3, 2013

tempneff

1. The problem statement, all variables and given/known data
A 100 nC point charge is located at A(-1, 1, 3) in free space. (a) Find the locus of all point P(x,y,z) at which $E_x = \frac{V}{m}.$ Find $y_1$ if P(-2,$y_1,3$) lies on that locus.
2. Relevant equations
$E=k\frac{q}{r^2}\vec{r}\hspace{10pt}E_X=E\cos\theta\hspace{10pt}\vec{R}=<(x_2-x_1),(y_2-y_1),(z_2-z_1)>\hspace{10pt}$
Cartesian to Spherical: $r=\sqrt{x^2+y^2+z^2}\hspace{10pt}\theta = cos^{-1}\frac{z}{\sqrt{x^2+y^2+z^2}} \hspace{10pt} \phi = \cot^{-1}\frac{x}{y}$

3. The attempt at a solution
I converted the vector R to spherical coordinates because I figured that using the radius I could find the field where $\theta$ and $\phi = 0$ then convert back to cartesian...It didn't work. I had a hint that I should use $E=k\frac{q}{\vert{\vec{R}\vert^3}}\vec{R}$ but I don't quite understand why. I believe this is just multiplying by 1, but I thought that while $k\frac{q}{r^2}\vec{r}$ already include the vector I could just plug in $\vec{R}$ for unit vector $\vec{r}$ then why would I have to divide by the magnitude again.

I am sure at this point that I am misunderstanding things so any clarification of these concepts it priceless. Thanks in advance for any tips.

2. Nov 3, 2013

haruspex

There seems to be a number missing. How many Volts/metre?
I see no merit in switching to polar. Use $E=k\frac{q}{r^2}\vec{r}$, replacing r with x, y, z. Given that you only want the x component, how will you replace $\vec{r}$?
(In the equation you posted with R3, the R vector appears to be the full vector, not the unit vector. Hence the need to divide by the magnitude again.)

3. Nov 3, 2013

tempneff

Right, that was supposed to be $E_x=500\frac{V}{m}$

I'm not sure how I can replace $\vec{r}$ reflecting only the x direction. Also, if they are asking for a locus, do i have to come up with a item including x,y, and z rather than just x.

4. Nov 3, 2013

haruspex

The complete vector ($\vec R$?) is <x, y, z>. The unit vector is the same divided by its magnitude. So what is the x component of that?
An equation involving x, y and z, yes.