Electric potential and Potential Energy of dipole Questions in English

(A) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} \right)$.
Since the charges are at the vertices of an equilateral triangle with side length $a$, we have $r_{12} = r_{23} = r_{31} = a$.
The charges are $q_1 = +q, q_2 = +2q, q_3 = +Q$.
Substituting these into the formula, we get: $U = \frac{k}{a} [ (q)(2q) + (2q)(Q) + (Q)(q) ]$.
Given that the potential energy $U = 0$, we have: $2q^2 + 2qQ + qQ = 0$.
$2q^2 + 3qQ = 0$.
$3qQ = -2q^2$.
$Q = -\frac{2q^2}{3q} = -\frac{2q}{3}$.

(B) The electrostatic potential energy $U$ of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.
For the given configuration, the charges are $Q$, $+2q$, and $+q$. The sides of the triangle are $\sqrt{2}a$, $\sqrt{2}a$, and the hypotenuse is $\sqrt{(\sqrt{2}a)^2 + (\sqrt{2}a)^2} = \sqrt{2a^2 + 2a^2} = \sqrt{4a^2} = 2a$.
The potential energy is $U = \frac{1}{4\pi\epsilon_0} [\frac{Q(2q)}{\sqrt{2}a} + \frac{Q(q)}{\sqrt{2}a} + \frac{(2q)(q)}{2a}] = 0$.
Multiplying by $4\pi\epsilon_0$ and simplifying:
$\frac{2Qq}{\sqrt{2}a} + \frac{Qq}{\sqrt{2}a} + \frac{2q^2}{2a} = 0$
$\frac{3Qq}{\sqrt{2}a} + \frac{q^2}{a} = 0$
$\frac{3Qq}{\sqrt{2}} = -q^2$
$Q = -\frac{\sqrt{2}q}{3}$.