Question

. Two fixed charges, +1.0 x 10 C and -3.0 x 106 C, are placed 10 cm apart. (a) Where maya third charge be located so that the net force acting on this charge is zero? (b) Is the equilibrium stable or not?

Answer 1

The charge $q_1=1*10^{-6}\,C$ is  placed at origin and charge $q_2=-3*10^{-6}\,C$ is at a distance of $d=10\,cm$ to the right side of charge 1.

A third charge must be placed at a distance $x$ to the left side of smaller charge $q_1=1*10^{-6}\,C$ , so that force $F_{31}$ on the third charge$q_3$ due to charge $q_1$  (directed towards left) is equal to the force $F_{32}$ on charge $q_3$ due to $q_2$ (directed towards right)

$F_{31}=\frac{kq_1q_3}{x^2}$$F_{32}=\frac{kq_2q_3}{(10+x)^2}$

Net force on charge $q_3$ is zero.

$F_{3}=-F_{31}+F_{32}=-\frac{kq_1q_3}{x^2}+\frac{kq_2q_3}{(10+x)^2}=0$

$\frac{kq_1q_3}{x^2}=\frac{kq_2q_3}{(10+x)^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(10+x)^2}$

${q_1}{(10+x)^2}=q_2\,x^2$

$1*10^{-6}{(10+x)^2}=3*10^{-6}\,(x)^2$

${(10+x)^2}=3\,(x)^2$

$-2x^2+20x+100=0$

$x=-3.66\,cm$

That is the third charge is placed on the negative X-axis at $x=-3.66\,cm$.

(Or) the third charge is placed at a distance of $3.66\,cm$ to the left side of charge $q_1$ and at a distance of $13.66\,cm$ to the left side of charge $q_2$.

(b)

The equilibrium is not stable.