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Question

(A) The electrical 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 \le

Vedclass · Accepted Answer

$q = -\frac{2}{3} Q$

Vedclass · Answer

(A) The electrical potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = k \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} ight)$.For an equilateral triangle with side length $a$,the distance between any two charges is $a$.Given charges are $q_1 = Q$,$q_2 = 2Q$,and $q_3 = q$.The total potential energy is $U = \frac{k}{a} (Q \cdot 2Q + 2Q \cdot q + q \cdot Q)$.Setting $U = 0$ for the system to have zero potential energy:$2Q^2 + 2Qq + qQ = 0$.$2Q^2 + 3Qq = 0$.$Q(2Q + 3q) = 0$.Since $Q 
eq 0$,we have $2Q + 3q = 0$.Therefore,$q = -\frac{2}{3} Q$.

Three point charges $+Q$,$+2Q$,and $q$ are placed at the vertices of an equilateral triangle. The value of charge $q$ in terms of $Q$,so that the electrical potential energy of the system is zero,is given by:

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