Locus of all points due to a charge

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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.
 

Answers and Replies

  • #2
haruspex
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[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.)
 
  • #3
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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.
 
  • #4
haruspex
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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.
 

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