Two point charges of $q_1=4\times 10^{-6}\text{ C}$ and $q_2=-2\times 10^{-6}\text{ C}$ are separated by a distance of $1\text{ m}$ in vacuum. At what distance from the positive charge on the line joining the two charges is the electric potential zero?

Solution

1. Let $x$ be the distance from the positive charge $q_1$ where the potential is zero, meaning the distance from $q_2$ is $1 - x$.
2. The total electric potential $V$ is given by $$\begin{array}{rl}V& =\frac{k{q}_1}{x}+\frac{k{q}_2}{1-x}\\ & =0\end{array}$$.
3. Substituting the given values, we have $\frac{4\times 10^{-6}}{x}+\frac{-2\times 10^{-6}}{1-x}=0$.
4. Simplifying the equation gives $\frac{4}{x}=\frac{2}{1-x}$, which leads to $4 - 4x = 2x$.
5. Solving for $x$ yields $6x = 4$, so $x=\frac{2}{3}\approx 0.67\text{ m}$.
6. Hence the answer is (C).