- #1

- 2,580

- 1

## Homework Statement

[PLAIN]http://img339.imageshack.us/img339/2158/71254129.jpg [Broken]

In the figure above, determine the point (other than infinity) at which the electric field is zero. Given q

_{1}= -2.50[tex]\mu[/tex]C and q

_{2}= 6.00[tex]\mu[/tex]C

Solution: 1.82m to the left of q

_{1}

## The Attempt at a Solution

So I set it up as

[tex]\frac{q_{1}}{x^2} = \frac{q_{2}}{(d-x)^2}[/tex] where is x is the point I am looking for

After some simplification

[tex]q_{1}d^2 - 2dq_{1}x + x^2(q_{1} - q{2}) = 0 [/tex]

Here is what hit me, if I was only going to concern about the magnitude, I would get x = -1.82m and 0.39m

First of all, I am not even sure why it is -1.82m and not 0.39. Isn't -1.82m outside the range of d = 1.00m?

Secondly, if I do not concern with the magnitude (using -2.50µC instead of 2.50µC), I couldn't even solve the quadratic.

So why must I use positive 2.50µC? And why is it -1.82m and not 0.39m?

Thanks!

Last edited by a moderator: