Computing Electric Field From The Potential

[faint545]

Three identical point charges, each with a charge equal to q, lie in the xy plane. Two of the charges are on the y axis at y = -a and y = +a, and the third charge is on the x axis at x = a.

a) Find the potential as a function of position along the x axis.
b) Use the part a) result to obtain an expression for $E$_x(x), the x component of the electric field as a function of x. Check your answers to parts a) and b) at the origin and as x approaches $\infty$ to see if they yield the expected results.

Hi all. I'm having a very hard time understanding this portion of Physics so please bear with me.

The furthest I got with this problem is deciding to use the sum of the potentials at each point to calculate the potential of the system. Something like...

$\frac{kq_{1}}{r_{1}}$ + $\frac{kq_{2}}{r_{2}}$ + $\frac{kq_{3}}{r_{3}}$


I think that's the right approach since they are point charges. But now, I'm completely stuck and I don't know what I should do next... Please help!

 

[SammyS]

What are the values of r₁, r₂, and r₃ at position x along the x-axis ?

 

[faint545]

What are the values of r₁, r₂, and r₃ at position x along the x-axis ?

Here's what I've done in regards to your question... I hope I am on the right track...

link: http://dl.dropbox.com/u/244748/2011-09-23%2018.29.26.jpg

 

[SammyS]

$\sqrt{x^2+a^2}\ne x+a$

However, $\sqrt{x^2+(-a)^2}=\sqrt{x^2+a^2}\,.$

r₃ = | x - a | . If x > a then |x - a| = x -a . Otherwise, |x - a| = a - x .

So, V(x) = ?

 

[faint545]

$\sqrt{x^2+a^2}\ne x+a$

However, $\sqrt{x^2+(-a)^2}=\sqrt{x^2+a^2}\,.$

r₃ = | x - a | . If x > a then |x - a| = x -a . Otherwise, |x - a| = a - x .

So, V(x) = ?

So I suppose $V(x) = kq(\frac{2}{\sqrt{x^{2}+a^{2}}} + \frac{1}{a-x})$ if $x < a$ and $V(x) = kq(\frac{2}{\sqrt{x^{2}+a^{2}}} + \frac{1}{x-a})$ if $x > a$.

I think I can compute $E_{x}(x)$ from here.. Thanks for all your help!

 

[SammyS]

Looks good !