Locus of all points due to a charge

[tempneff]

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 [itex]E_x = \frac{V}{m}.[/itex] Find [itex]y_1[/itex] if P(-2,[itex]y_1,3[/itex]) lies on that locus.

Homework Equations

[itex]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}[/itex]
Cartesian to Spherical: [itex]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}[/itex]

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 [itex]\theta[/itex] and [itex]\phi = 0[/itex] then convert back to cartesian...It didn't work. I had a hint that I should use [itex]E=k\frac{q}{\vert{\vec{R}\vert^3}}\vec{R}[/itex] but I don't quite understand why. I believe this is just multiplying by 1, but I thought that while [itex]k\frac{q}{r^2}\vec{r}[/itex] already include the vector I could just plug in [itex]\vec{R}[/itex] for unit vector [itex]\vec{r}[/itex] 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]

[itex]E_x = \frac{V}{m}.[/itex]

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.)

 

[tempneff]

Right, that was supposed to be [itex]E_x=500\frac{V}{m}[/itex]

I'm not sure how I can replace [itex]\vec{r}[/itex] 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 [itex]\vec{r}[/itex] reflecting only the x direction.

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?

Also, if they are asking for a locus, do i have to come up with a item including x,y, and z

An equation involving x, y and z, yes.

 

[Nickie]

The locus of all points due to a charge is the set of points where the electric field has a magnitude of V/m, as given in the problem. The equation for electric field is E = kq/r^2, where k is the Coulomb constant, q is the charge, and r is the distance from the charge.

In this problem, we are given a point charge of 100 nC located at (-1,1,3) and are asked to find the locus of points where E_x = V/m. This means that the electric field in the x-direction (E_x) at these points must equal V/m.

To find this locus, we can use the equation E_x = kq/r^2 * cos(theta), where theta is the angle between the direction of the electric field and the x-axis. We can rewrite this equation as E_x = kq/r^3 * x, where x is the distance in the x-direction from the charge.

Since we know that E_x = V/m and q = 100 nC, we can solve for r to find the locus of points. This gives us r^3 = kq/Vx, and substituting in the values gives us r = (kq/Vx)^(1/3).

To find y_1, we can use the given point P(-2,y_1,3) and plug in the values for r and x. This gives us r = (kq/V(-2))^(1/3) = (-kq/2V)^(1/3). Since we know that r = sqrt(x^2+y^2+z^2), we can solve for y_1 by rearranging the equation to y_1 = sqrt(r^2-x^2-z^2) = sqrt((kq/2V)^(2/3) - 4).

Therefore, the locus of points where E_x = V/m is given by r = (kq/Vx)^(1/3), and the value of y_1 at P(-2,y_1,3) is given by y_1 = sqrt((kq/2V)^(2/3) - 4).

Note: The hint you were given to use the equation E = kq/r^3 * R is because we are finding the locus of points where E_x = V/m, which means that the