# Locus of all points due to a charge

## Homework Statement

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.

## Homework 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: $r=\sqrt{x^2+y^2+z^2}\hspace{10pt}\theta = cos^{-1}\frac{z}{\sqrt{x^2+y^2+z^2}} \hspace{10pt} \phi = \cot^{-1}\frac{x}{y}$

## 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.

haruspex
Homework Helper
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2020 Award
$E_x = \frac{V}{m}.$
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.)

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.

haruspex
I'm not sure how I can replace $\vec{r}$ reflecting only the x direction.