Homework Help: Electric field - point of zero field

1. Sep 14, 2010

Uku

1. The problem statement, all variables and given/known data
Find a point on the line connecting two charges $$q_{1}$$ and $$q_{1}$$
where the electric field strength is zero. These charges $$q_{1}$$ and $$q_{1}$$ at a distance of $$d$$.
NB - the charges can be of the same charge or different charge.

3. The attempt at a solution
I know the answer to be (say x is the distance):

$$x=\frac{\sqrt{q_{1}}}{\sqrt{q_{1}}\pm\sqrt{q_{2}}}d$$ (1)

Right, for starters I assume that these charges are both positive? and I place a negative test charge to the right of the two charges, at a distance of $$x$$. This means that the test charge is at a distance $$d+x$$ from the first charge.
Right, without revealing the in-between I reach a point where, by superposition

$$E_{1}-E_{2}=0$$

and dividing that by k

$$\frac{q_{1}}{(d+x)^{2}}-\frac{q_{2}}{x^{2}}=0$$

from where I reach the following by taking a square root of the above expression (by which I lose one solution?)

$$x=\frac{\sqrt{q_{2}}}{\sqrt{q_{1}}+\sqrt{q_{2}}}d$$

but that does not match up with the answer in the book. (1)
Especially the fact that $$q_{2}$$ is in the top of the division, not $$q_{1}$$, the plus-minus part I don't mind.

2. Sep 14, 2010

rl.bhat

Right, for starters I assume that these charges are both positive?

If both the charges are positive, the point of zero electric field will in between the two charges, i.e. if x is the distance from one charge, d-x will be the distance from the other charge. The value of x depends on the charge from which the distance is measured.

3. Sep 14, 2010

Uku

If the charges are positive yes, even simply by using a mind experiment, the zero point is in between them. Depending on the charge of either $$q_{1}$$ or $$q_{2}$$, the distance of the zero point from the first charge (say it is the starting point of the coordinate system) will vary by d-x...

And I don't know where I went wrong... I actually did the calculation before, but I assumed $$q_{1}=q_{2}$$ and that gave me $$x= \frac{d}{2}$$, which makes sense when keeping my assumption in mind.

Thanks!

4. Sep 14, 2010

rl.bhat

At neutral point

q1/x^2 = q2/(d-x)^2

Simplify and find the value of x.