Question

Two fixed charges 1.07 micro Coulomb and -3.29micro Coulomb are 61.8cm apart where may a third charge be located so that no net force act on it?

need better explantion please

Answer 1

The third charge should be placed at the point, where two electric fields due to charges compensate each other (equal). Let $d=61.8cm = 0.618m$ be the distance between charges, $q_1 = 1.07\times 10^{-6}C, q_2 = -3.29\times 10^{-6}C$ charges, $x$ - distance from the first charge. Then, at the point of zero resultant force have:

$$k\dfrac{|q_1|}{x^2}=k\dfrac{|q_2|}{(d+x)^2}$$

The equation for $x$ is then:

$$|q_1|(d^2+2dx+x^2) = |q_2|x^2\\ (|q_1|-|q_2|)x^2+2d|q_1|x + |q_1|d^2=0$$

Substituting numbers and solving the quadratic equation for $x$, obtain:

$$x\approx 0.82m$$

Answer. The third charge can be located at 0.82m from the first one.