The value of b is ___ meter.

The electric potential at any point $$P(x,y)$$ in the xy-plane due to the two charges is

$$V = k \left(\frac{-Q}{r_1} + \frac{Q/\sqrt{3}}{r_2}\right)$$
where $$r_1 = \sqrt{x^2 + y^2}$$ is the distance from $$P$$ to the charge $$-Q$$ at the origin, and $$r_2 = \sqrt{(x-2)^2 + y^2}$$ is the distance from $$P$$ to the charge $$+\dfrac{Q}{\sqrt{3}}$$ at $$(2,0)$$.

An equipotential circle of potential $$V = 0$$ satisfies

$$\frac{-Q}{r_1} + \frac{Q/\sqrt{3}}{r_2} = 0 \;\Longrightarrow\; \frac{1}{r_1} = \frac{1}{\sqrt{3}\,r_2} \;\Longrightarrow\; r_1 = \sqrt{3}\,r_2 \; -(1)$$

Thus every point on the required circle has its distance from the origin $$\sqrt{3}$$ times its distance from $$(2,0)$$. The locus of points whose distances from two fixed points are in a constant ratio is an Apollonius circle.

Let $$m = \sqrt{3}$$. Using coordinates, the condition $$r_1 = m\,r_2$$ becomes

$$x^2 + y^2 = m^{2}\Bigl((x-2)^2 + y^2\Bigr)$$

Substituting $$m^{2}=3$$ gives

$$x^2 + y^2 = 3\bigl((x-2)^2 + y^2\bigr) \;\Longrightarrow\; x^2 + y^2 = 3(x^2 - 4x + 4 + y^2)$$

$$\Longrightarrow\; x^2 + y^2 = 3x^2 - 12x + 12 + 3y^2$$

Bringing all terms to one side and dividing by 2:

$$0 = 2x^2 - 12x + 12 + 2y^2 \;\Longrightarrow\; x^2 - 6x + y^2 + 6 = 0$$

Complete the square in $$x$$:

$$(x-3)^2 - 9 + y^2 + 6 = 0 \;\Longrightarrow\; (x-3)^2 + y^2 = 3$$

This is the equation of a circle with
center $$(b,0) = (3,0)$$ and radius $$R = \sqrt{3}\;\text{m}$$.

Therefore,

$$b = 3.00\;\text{m}$$

The value falls inside the accepted range $$2.90 - 3.10$$.