Step 1: Understanding the Concept:
The electrical potential energy of a system of point charges is the sum of the potential energies of all unique pairs of charges in the system.
Step 2: Key Formula or Approach:
For a system of three charges $q_1, q_2, q_3$ separated by distances $r_{12}, r_{23}, r_{13}$, the total electrostatic potential energy $U$ is:
\[ U = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right) \]
For an equilateral triangle, all side lengths are equal. Let the side length be '$a$'.
Step 3: Detailed Explanation:
Let the three charges be $q_1 = +Q$, $q_2 = +2Q$, and $q_3 = q$.
The distances are $r_{12} = r_{23} = r_{13} = a$.
Substitute the charges into the potential energy formula:
\[ U = \frac{1}{4\pi\varepsilon_0} \left( \frac{(Q)(2Q)}{a} + \frac{(2Q)(q)}{a} + \frac{(Q)(q)}{a} \right) \]
We are given that the total potential energy of the system is zero ($U = 0$):
\[ 0 = \frac{1}{4\pi\varepsilon_0 a} \left( 2Q^2 + 2Qq + Qq \right) \]
Since $\frac{1}{4\pi\varepsilon_0 a}$ is a non-zero constant, the sum inside the parenthesis must be zero:
\[ 2Q^2 + 3Qq = 0 \]
Assuming $Q \neq 0$, we can divide the entire equation by $Q$:
\[ 2Q + 3q = 0 \]
Now, solve for $q$ in terms of $Q$:
\[ 3q = -2Q \]
\[ q = -\frac{2}{3}Q \]
Step 4: Final Answer:
The value of charge $q$ is $-\frac{2}{3}\text{Q}$.