Three charges Q0,-q and -q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q0 is equal to

Question

Three charges Q0,-q and -q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q0 is equal to (A)  (q/4)  (B)  (2q/√32)  (C)  √2q  (D)  +q

Tardigrade Admin · Accepted Answer

Here total electrostatic potential energy is zero ie, (-Q0q/l)-(qQ0/l)+(q2/√2l)=0 On solving, Q0=(q/2√2)=(2q/√32)

Q. Three charges $Q_{0}, - q$ and $- q$ are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if $Q_{0}$ is equal to

$\frac{q}{4}$

$\frac{2 q}{32}$

$2 q$

$+ q$

$\frac{q}{2}$

Here total electrostatic potential energy is zero ie,
$\frac{- Q _{0} q}{l} - \frac{q Q _{0}}{l} + \frac{q ^{2}}{2 l} = 0$
On solving, $Q_{0} = \frac{q}{2 2} = \frac{2 q}{32}$

Q. Three charges Q0​,−q and −q are placed at the vertices of an isosceles right angle triangle as in the figure. The net electrostatic potential energy is zero if Q0​ is equal to

4q​

32​2q​

2q​

+q

2​q​

Here total electrostatic potential energy is zero ie, l−Q0​q​−lqQ0​​+2​lq2​=0 On solving, Q0​=22​q​=32​2q​