Question

A charge of 1.98E-9 C is placed at the origin, and a charge of 3.76E-9 C is placed at x = 1.37 m. Find the position at which a third charge of 2.88E-9 C can be placed so that the net electrostatic force on it is zero.

Answer 1

We can use electric field equations

so

$\frac{k\times q_1}{(1.37+x)^2} = \frac{k\times q_2}{x^2}$

$1.98\times 10^{-9}\times x^2} =3.76\times 10^{-9}\times{(1.37+x)^2}$

By solving we get

x= -0.7938 m or x=-4.9990 m