# Homework Help: Electric Potential and Potential Energy Due to Charged Particles

1. Apr 2, 2013

### brojas7

Two charged particles of equal magnitude are located along the y axis equal distances above and below the x-axis.
Plot a graph of the electric potential at points along the x axis over the interval -3a<x<3a. You should plot the potential in units of (KeQ)/a
Let the charge of the particle located at y=-a be negative. Plot the potential along the y axis over the interval -4a<y<4a.

I was able to do :
(KeQ1)/r^2 + (KeQ2)/r^2 =
(KeQ)/ sqrt(x^2 + a^2) + (KeQ)/ sqrt(x^2 + -a^2) =
(2KeQ)/ sqrt(x^2 + -a^2)

Why did they factor out a KeQ/a, was it necessary? how do you plug in x to get the graph if you don't know 'a'.

and for the y axis, why did they get an absolute value for y-a and y+a.

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2. Apr 2, 2013

### rude man

It made it possible to graph V as a function of x/a only. You don't need to know a if you're happy with x/a for the x axis (really making it an "x/a" axis).
Because r is always positive, so the quantity y-a or y- (-a) must be positive and therefore put in absolute bars. y can be as negative as -4a, remember. (a is always positive.)

BTW your 1st eq. has a typo in it. r, not r^2, as you know.

3. Apr 2, 2013

### brojas7

Ok the first part makes sense as for the second, since r is always positive and must put in the absolute bars, why didn't we do that with the first? Is it because the square root already takes care of that?

4. Apr 2, 2013

### brojas7

Also, why must r always be positive? Or is this just a rule?
And thank you for the correction, I completely overlooked it, even on paper.

5. Apr 2, 2013

### rude man

Right. You just ignore the negative root.

6. Apr 2, 2013

### rude man

You can't have negative distance! The sign of the potential due to a charge is determined solely by the sign of the charge.

7. Apr 2, 2013

### brojas7

Ok it all makes sense now. I didn't think of it as a distance. thank you so much