Problem 19.9

19.9.
From symmetry considerations, we realize that the point of zero electric field must lie along the line connecting the two charges. Also, since the negative charge is greater in magnitude than the positive one, the point of zero electric field must lie closer to the negative charge than the positive one. Let's call the distance from the point we are looking for to the negative charge d. Then the distance between the point of zero field and the positive charge is d + 1m. Let's calculate the electric field due to both charges at our point.
Let the electric field due to the negative charge be given by E_- and the electrice field due to the positive charge be given by E₊.
$\vec{E_{-}} = \frac{kq}{r} \hat{r}$
$\vec{E_{-}} = \frac{(k)(-2.5 \times 10^{-6} C)}{d^{2}} (-\hat{r})$
Note that the point at d is to the left of the charge on the axis, so the direction vector has a negative sign.
$\vec{E_{+}} = \frac{(k)(6 \times 10^{-6} C)}{(d+1)^{2}} (-\hat{r})$
Where we are now 1 m further away from the charge. Now, if we add the contributions from E_- and E₊, we should get zero field, or put another way,
E_- = -E₊
Simplifying yields
(d+1)² = 2.4 d²
Solving this with the quadratic equation we get two values for d.
d = -.392 m, 1.82 m
We now must examine the problem to determine which of these is the correct answer. What we actually found was the two locations where the electric fields from each charge have equal magnitudes. However, the soltuion d = -.392 m can not be the answer because even though the contributions to the electric field from both charges have equal magnitudes, they have the same direction, so therefore they would add to give a nonzero value for E. The correct answer where the two components have equal magnitudes and opposite directions is
d = 1.82 m to the left of the $-2.5 \times 10^{-6} C$ charge.