Three point charges +q, +2q and +Q are placed | vedclass

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(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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: $2q^2 + 2qQ + qQ = 0$.$2q^2 + 3qQ = 0$.$3qQ = -2q^2$.$Q = -\frac{2q^2}{3q} = -\frac{2q}{3}$.

Three point charges $+q, +2q$ and $+Q$ are placed at the vertices of an equilateral triangle. If the potential energy of the system of three charges is zero, the value of $Q$ in terms of $q$ is:

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Three point charges $+q$, $+2q$, and $+Q$ are placed at the three vertices of an equilateral triangle. If the potential energy of the system of three charges is zero, the value of $Q$ in terms of $q$ is

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