Question

Given that q=+12μC, find the distance from q1 where q2 experiences a net electrostatic force of zero. (The charges q1 and q3 are separated by a fixed distance of 42 cm .)

Answer 1

Missing information in the text of the problem: q3=3 q1 = 3q

We can solve the problem by solving a system of two equations. In the first one, we write that the electrostatic force between charge 1 and 2 is equal to the force between charge 2 and 3. In the second equation, we write that the sum of the distances q1-q2 and q2-q3 is equal to 42 cm:
$k_e \frac{q_1 q_2}{r_{12}^2}=k_e \frac{q_2 q_3}{r_{23}^2}$
$r_{12}+r_{23}=0.42~m$
From the second equation we write
$r_{12}=0.42-r_{23}$
While we can rewrite the first equation as
$\frac{q}{r_{12}^2}= \frac{3q}{r_{23}^2}$
Substituting and solving, we find
$r_{23}=0.99~m$ or $r_{23}=0.27~m$. With the first solution, $r_{12}$ would be negative, which doesn't make sense; therefore we must consider only the second solution, and we have:
$r_{23}=0.27~m$
$r_{12}=0.42-r_{23}=0.15~m$