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

Let the point where the electric field is zero be at a distance $x$ from charge $A$.

Since the charges have opposite signs, the null point must lie outside the line segment joining the two charges, on the side of the charge with the smaller magnitude. Here, $|q_A| < |q_B|$, so the point lies on the side of $A$, which corresponds to the direction along $BA$.

At the null point, the magnitudes of the electric fields due to both charges must be equal:

$\dfrac{1}{4\pi\epsilon_0} \dfrac{|q_A|}{x^2} = \dfrac{1}{4\pi\epsilon_0} \dfrac{|q_B|}{(x + d)^2}$

Substituting the given values:

$\dfrac{2.00 \times 10^{-6}}{x^2} = \dfrac{4.00 \times 10^{-6}}{(x + 20.0)^2}$

$\dfrac{1}{x^2} = \dfrac{2}{(x + 20.0)^2}$

Taking the square root of both sides:

$\dfrac{1}{x} = \dfrac{\sqrt{2}}{x + 20.0}$

$x + 20.0 = \sqrt{2}x$

$x(\sqrt{2} - 1) = 20.0$

$x = \dfrac{20.0}{\sqrt{2} - 1}$

Rationalizing the denominator:

$x = 20.0(\sqrt{2} + 1)$

Using $\sqrt{2} \approx 1.414$:

$x = 20.0(1.414 + 1) = 20.0 \times 2.414 = 48.28 \text{ cm}$

Rounding off to three significant figures, the distance is $48.3 \text{ cm}$.

Thus, the electric field is zero at a distance of $48.3 \text{ cm}$ from $A$ along $BA$.