Step-by-Step Solution

1. Identify the setup and objective: We are given two point charges: $q_1 = -Q$ located at the origin (0, 0) and $q_2 = +Q/\sqrt{3}$ located at (2, 0). We need to find the radius R of an equipotential circle in the xy-plane where the electric potential V is zero.

2. Formulate the electric potential equation: Let P(x, y) be an arbitrary point on the equipotential circle. The electric potential at point P is the algebraic sum of the potentials due to charges $q_1$ and $q_2$.

   The distance from $q_1$ to P is $r_1 = \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$. The distance from $q_2$ to P is $r_2 = \sqrt{(x-2)^2 + (y-0)^2} = \sqrt{(x-2)^2 + y^2}$.

   The total potential V at P(x, y) is given by: $V = V_1 + V_2 = \frac{k q_1}{r_1} + \frac{k q_2}{r_2}$ Substituting the given values: $V = \frac{k(-Q)}{\sqrt{x^2 + y^2}} + \frac{k(+Q/\sqrt{3})}{\sqrt{(x-2)^2 + y^2}}$

3. Apply the equipotential condition (V=0): We are given that the potential on the circle is zero, so we set V = 0. $\frac{k(-Q)}{\sqrt{x^2 + y^2}} + \frac{k(Q/\sqrt{3})}{\sqrt{(x-2)^2 + y^2}} = 0$ Assuming k and Q are non-zero, we can simplify the equation: $\frac{Q/\sqrt{3}}{\sqrt{(x-2)^2 + y^2}} = \frac{Q}{\sqrt{x^2 + y^2}}$ $\frac{1}{\sqrt{3}\sqrt{(x-2)^2 + y^2}} = \frac{1}{\sqrt{x^2 + y^2}}$ This implies: $\sqrt{x^2 + y^2} = \sqrt{3} \sqrt{(x-2)^2 + y^2}$

4. Derive the equation of the circle: To find the locus of the point P(x, y), we square both sides of the equation to eliminate the square roots: $x^2 + y^2 = 3 \left[ (x-2)^2 + y^2 \right]$ Expand the terms: $x^2 + y^2 = 3 (x^2 - 4x + 4 + y^2)$ $x^2 + y^2 = 3x^2 - 12x + 12 + 3y^2$ Rearrange all terms to one side: $(3x^2 - x^2) + (3y^2 - y^2) - 12x + 12 = 0$ $2x^2 + 2y^2 - 12x + 12 = 0$ Divide the entire equation by 2: $x^2 + y^2 - 6x + 6 = 0$

5. Determine the radius of the circle: To find the radius, we convert the equation to the standard form of a circle, $(x-b)^2 + (y-k)^2 = R^2$, by completing the square for the x terms. $(x^2 - 6x) + y^2 = -6$ To complete the square for x, we add $(6/2)^2 = 3^2 = 9$ to both sides: $(x^2 - 6x + 9) + y^2 = -6 + 9$ $(x - 3)^2 + y^2 = 3$ Comparing this with the standard form, we have the center at (b, 0) = (3, 0) and the radius squared $R^2 = 3$. Therefore, the radius of the equipotential circle is: $R = \sqrt{3} \text{ meters}$

6. Calculate the final numerical value: The numerical value of the radius is $R = \sqrt{3} \approx 1.73205...$ meters. Rounding to two decimal places, we get R = 1.73 meters.