Three charges Q,(.+q.) and (.+q.) are placed at the vertices of an equilateral triangle of side 1m as shown in the figure. If the net electrostatic energy of the system is zero, then Q is equal to:

Question

Three charges Q,(.+q.) and (.+q.) are placed at the vertices of an equilateral triangle of side 1m as shown in the figure. If the net electrostatic energy of the system is zero, then Q is equal to: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-8whyz7cg5fb7ljny.jpg" /> (A) (- (q/2)) (B) (- q) (C) (+ q) (D) Zero

Tardigrade Admin · Accepted Answer

U=(k q Q/l)+(k q Q/l)+(k q2/l)=0 Q=-(q/2)

Q. Three charges $Q, (+ q)$ and $(+ q)$ are placed at the vertices of an equilateral triangle of side $1 m$ as shown in the figure. If the net electrostatic energy of the system is zero, then $Q$ is equal to:

$(- \frac{q}{2})$

$(- q)$

$(+ q)$

Zero

$U = \frac{k qQ}{l} + \frac{k qQ}{l} + \frac{k q ^{2}}{l} = 0$
$Q = - \frac{q}{2}$

Q. Three charges Q,(+q) and (+q) are placed at the vertices of an equilateral triangle of side 1m as shown in the figure. If the net electrostatic energy of the system is zero, then Q is equal to:

(−2q​)

(−q)

(+q)