Answer to Question #124614 in Electricity and Magnetism for Kenneth

Question

Two points charges are placed on the x-axis: a – 6 µC charge at x = 0 and a + 5µC charge at x = 200 cm. At
what point(s) in the vicinity of these two charges can a + 5- µC charge be placed to that it experiences no
resultant force?

Answer 1

Let x be the coordinate of the third charge. The force from the first charge is

$\vec{F_1} = k\dfrac{q_1q_1}{r_{12}^3}\vec{r}_{12}$ , $|F_1| = k \cdot\dfrac{5\,\mathrm{\mu C}\cdot 6\,\mathrm{\mu C}}{(x-0)^2} = k\cdot\dfrac{30}{x^2}$ and this force is directed from x to 0.

The force from the second charge is

$|F_2| = k \cdot\dfrac{5\,\mathrm{\mu C}\cdot 5\,\mathrm{\mu C}}{(x-2)^2} = k\cdot\dfrac{25}{(x-2)^2}$ and this force is directed from 2m to x.

Let the charge be placed between two charges. The net force is 0, so

$\dfrac{-30k}{x^3}x - \dfrac{25k}{(x-2)^3}(x-2) = 0. \;\; \dfrac{-30k}{x^2} - \dfrac{25k}{(x-2)^2} = 0.$ $\;\;\; \dfrac{6}{x^2} =- \dfrac{5}{(x-2)^2}.$

$11x^2-24x+24 = 0.$ This equation has no roots.

Let the charge be placed left to the first charge, so

$\dfrac{30k}{x^2} - \dfrac{25k}{(x-2)^2} = 0. \;\;\; \dfrac{6}{x^2} =\dfrac{5}{(x-2)^2}.$ $x^2-24x+24 = 0. \;\; x = 12 \pm2\sqrt{30}$ , but none of these roots is less than 0.

Let the charge be placed right to the second charge, so

$-\dfrac{30k}{x^2} -+\dfrac{25k}{(x-2)^2} = 0. \;\;\; \dfrac{6}{x^2} =\dfrac{5}{(x-2)^2}.$ $x^2-24x+24 = 0. \;\; x = 12 \pm2\sqrt{30}$ . We choose the root $x = 12+2\sqrt{30} \approx 22.95\,\mathrm{m}.$