Determine Center Position of Equipotential Circle

To find the value of $b$, we analyze the electric potential in the $xy$-plane created by the two point charges.

The electric potential $V(x, y)$ at any point $(x, y)$ in the $xy$-plane due to the charges $q_1 = -Q$ at $(0, 0)$ and $q_2 = +\frac{Q}{\sqrt{3}}$ at $(2, 0)$ is given by the principle of superposition:

$V(x, y) = \frac{1}{4\pi\varepsilon_0} \left( \frac{-Q}{\sqrt{x^2 + y^2}} + \frac{\frac{Q}{\sqrt{3}}}{\sqrt{(x - 2)^2 + y^2}} \right)$

We are given that the equipotential surface corresponds to $V(x, y) = 0$. Setting the total potential to zero:

$\frac{1}{4\pi\varepsilon_0} \left( \frac{-Q}{\sqrt{x^2 + y^2}} + \frac{Q}{\sqrt{3}\sqrt{(x - 2)^2 + y^2}} \right) = 0$

Assuming $Q \neq 0$, we can simplify the equation as follows:

$\frac{1}{\sqrt{x^2 + y^2}} = \frac{1}{\sqrt{3}\sqrt{(x - 2)^2 + y^2}}$

Squaring both sides to remove the square roots:

$\frac{1}{x^2 + y^2} = \frac{1}{3\left[(x - 2)^2 + y^2\right]}$

Cross-multiplying gives:

$3\left[(x - 2)^2 + y^2\right] = x^2 + y^2$

Expanding the term $(x - 2)^2$:

$3(x^2 - 4x + 4 + y^2) = x^2 + y^2$

$3x^2 - 12x + 12 + 3y^2 = x^2 + y^2$

Rearranging all terms to one side:

$2x^2 - 12x + 2y^2 + 12 = 0$

Dividing the entire equation by $2$:

$x^2 - 6x + y^2 + 6 = 0$

Now, completing the square for the $x$-terms:

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

$(x - 3)^2 + y^2 = 3$

This is the equation of a circle of the standard form $(x - b)^2 + y^2 = R^2$, where:

• The center of the circle is at $(b, 0) = (3, 0)$
• The radius of the circle is $R = \sqrt{3}\text{ m}$

Therefore, the value of $b$ is $3$.