Want to know correct answer for electric field + potential question

  • Thread starter altegron
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  • #1
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Homework Statement



http://img352.imageshack.us/img352/9444/enm1qi8.png [Broken]

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



[tex]V = {k}_{e}\frac{q}{r}[/tex]

[tex]E = {k}_{e}\frac{q}{r^2}[/tex]

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:

[tex]0 = 2({k}_{e}\frac{q}{\sqrt{a^2+d^2}}) - {k}_{e}\frac{q}{d}
[/tex]

And simplify/solve for d to get:

[tex]d = \frac{\sqrt{3}a}{3}[/tex]

(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:

[tex]E = 2({k}_{e}\frac{q}{(\sqrt{a^2+x^2})^2})[/tex]

and to get the x component multiplied that by:

[tex]{E}_{x} = \cos\tan^{-1}\frac{a}{x}[/tex]

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

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

[tex]{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}[/tex]


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.
 
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Answers and Replies

  • #2
Shooting Star
Homework Helper
1,977
4
3a) d should have two values.

3b) What happened to the field due to the charge (-q)?
 
  • #3
14
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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 [itex]-{k}_{e}\frac{q}_{x^2}[/itex]?
 
  • #4
Shooting Star
Homework Helper
1,977
4
and to get the x component multiplied that by:

[tex]{E}_{x} = \cos\tan^{-1}\frac{a}{x}[/tex]

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

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

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

That is correct now.
 
  • #5
19
0
[tex]{E}_{x net} = 2(\cos\tan^{-1}\frac{a}{x})({k}_{e}\frac{q}{a^2+x^2}) - {k}_{e}\frac{q}_{x^2}[/tex]

can't you just replace [tex](\cos\tan^{-1}\frac{a}{x})[/tex] with [tex]\frac{x}{\sqrt{(x^2+a^2)}}[/tex]


...or is that what shootingstar was saying
 
Last edited:

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