# Electric potential

1. May 14, 2006

### Brewer

Sorry to be a pain again, but I'm struggling with my whole electromagnetic course, that I'm being examined on in a few days!

Next question I think may just be a problem with arithmetic and algebra, but we'll probably see that its more than that!

Question says:

A distribution of three charges is given below. Show that the electric potential at point P on the x-axis, where x>a is given by:
$$V(x) = \frac{x^2 + a^2}{x(x^2 - a^2)}$$​

Now I tried this. My plan was to work out the electric potential from each point and then add them up (linear superposition right?). Having been given that $$V(r) = \frac{q}{4\pi \epsilon_{0}r}$$ earlier I proceeded to use this.

Am I correct in thinking that the distance from q1 (the left hand charge) is 2a+x, the distance from q2 (center) is a+x and from q3 is just x?

Then when I added all these together I ended up still having the q/4pi etc in the fraction as well as another fraction that I got as $$\frac{2a-2x}{x(2a+x)}$$ instead of the required fraction.

Does anybody see where I went wrong with this?

Thanks again guys

2. May 14, 2006

### Staff: Mentor

No. x is just the x-coordinate of point P, measured from the origin. The distance between P and q1, for example, will be a + x, not 2a + x; the distance between P and q2 is just x; etc. (At least that's how I interpret the problem statement.)

3. May 14, 2006

### Brewer

Of course! So for q3 it will be x-a?

Does the rest of the way I was attempting the question look all right though?

4. May 14, 2006

### Staff: Mentor

Right.
Yes, adding the three potential terms is the way to go.

5. May 14, 2006

### Brewer

Now I end up with a constant term of $$\frac{q}{4\pi\epsilon_{0}}$$ outside of a bracket and $$\frac{2a^2}{x(x^2 - a^2)}$$ within a bracket. I need to get rid of the outside term (q/4pi etc) and I must have gone wrong with my arithmetic.

Any ideas?

6. May 14, 2006

### arunbg

Are you sure there is no $\frac{q}{4\pi\epsilon_{0}}$ in the question. Then the question is wrong. Also check your arithmetic.

7. May 14, 2006

### Brewer

Yep, very sure.

I've been through a couple of times now, and I end up with the $$2a^2$$ on the top everytime.

I'm rapidly losing faith in electromagnetism. My textbook and notes are no help either. :(

8. May 14, 2006

### Staff: Mentor

I don't see anything wrong with your solution. (Something's obviously wrong with the book's answer if it leaves out the $\frac{q}{4\pi\epsilon_{0}}$ term!)

9. May 14, 2006

### Brewer

Have I got my arithmetic wrong?

I keep trying, but I still end up with $$\frac{2a^2}{x(x^2 - a^2)}$$

10. May 14, 2006

### Staff: Mentor

As I said before, your answer is perfectly correct. The book is wrong. (Don't waste any more time stewing over this one. Move on to other problems. )