Two point charges of $q_1=2\mu \text{C}$ and $q_2=8\mu \text{C}$ are separated by a distance of $d=6\text{cm}$. The position of the neutral point where the electric field is zero is at a distance from the $2\mu \text{C}$ charge equal to:

Solution

1. Let the distance from the $2\mu \text{C}$ charge be $x$.
2. Equate the electric fields: $\frac{k{q}_1}{x^2}=\frac{k{q}_2}{(d-x{)}^2}$.
3. Substitute values: $\frac{2}{x^2}=\frac{8}{(6-x{)}^2}$.
4. Taking the square root: $$\begin{array}{rl}\frac{\sqrt{2}}{x}& =\frac{\sqrt{8}}{6-x}⟹\frac{1}{x}\\ & =\frac{2}{6-x}\end{array}$$.
5. Solving for $x$ gives $6-x = 2x$, so $x=2\text{cm}$.
6. Hence the answer is (B).