Question

A point charge $q_1=2\mu \text{C}$ is placed at the origin of coordinates. A second charge, $q_2=-3\mu \text{C}$, is placed on the $x-$ axis at $x=100\text{cm}$. For what positive $x-$coordinate of a point on the $x-$axis, the absolute potential at it is zero?

• A. $10\text{cm}$
• B. $20\text{cm}$
• C. $30\text{cm}$
• D. $40\text{cm}$

Answer 1

The correct option is D $40\text{cm}$

From system of charges and distances of point from charges, we can think that the zero potential will lie in between the given charges for positive $x-$coordinate. $\because V\propto Q\&V\propto \frac{1}{r}$


Thus the position of point should be near to the smaller $+ve$ charge so that its contribution in net potential increases.
Let the $x-$coordinate be $"a"$ then,
$V_{\text{net}}=0$
$V_{q1}+V_{q2}=0$
$\Rightarrow \frac{k{q}_1}{a}+\frac{k{q}_2}{1-a}=0$
$\Rightarrow \frac{2\times {10}^{-6}}{a}+\frac{(-3\times {10}^{-6})}{1-a}=0$
$\Rightarrow \frac{2}{a}=\frac{3}{1-a}$
$\Rightarrow 2-2a=3a$
$\Rightarrow a=0.4\text{m}=40\text{cm}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept: For the system of two charges, the net potential}\\ \text{at any point will be zero only when the two charges are of opposite nature.}\\ \text{Also, for different magnitude of charges, point of zero potential}\\ \text{will lie near to smaller charge.}\end{array}$