Two particles A and B , possessing charges of +2.00 × 10^ -6 C and...

Let $A$ be at the origin and $B$ be at a distance of $20 \text{ cm}$ from $A$.

The electric potential at a point $P$ is given by V = \dfrac{1}{4\pi\epsilon_0} \left(\dfrac{q_A}{r_A} + \dfrac{q_B}{r_B} \right).

For $V = 0$, we have \dfrac{q_A}{r_A} = -\dfrac{q_B}{r_B}.

Substituting the given charges:

\dfrac{2.00 \times 10^{-6}}{r_A} = \dfrac{4.00 \times 10^{-6}}{r_B} \Rightarrow$r_B = 2r_A$

Case 1: $P$ lies between $A$ and $B$.

$r_A + r_B = 20$

$r_A + 2r_A = 20 \Rightarrow 3r_A = 20 \Rightarrow r_A =$\dfrac{20}{3} \text{ cm}

This point is \dfrac{20}{3} \text{ cm} from $A$ along $AB$.

Case 2: $P$ lies outside the segment $AB$. Since $|q_A| < |q_B|$, the point must be closer to $A$, meaning it lies on the side of $A$ along the extended line $BA$.

$r_B - r_A = 20$

$2r_A - r_A = 20 \Rightarrow r_A = 20 \text{ cm}$

This point is $20 \text{ cm}$ from $A$ along $BA$.