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 <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-zyt3mvhnxjv2il62.jpg" /> (A) (q/4) (B) (q/√8) (C) √2q (D) q

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-zyt3mvhnxjv2jds6.jpg" /> U=(K Q0 (- q)/l)× 2+(K (- q) (- q)/l √2)=0 (K q/l)[- 2 Q0 + (q/√2)]=0 ∴ Q0=(q/2 √2)=(q/√8)

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

$\frac{q}{4}$

$\frac{q}{8}$

$2 q$

$q$

$U = \frac{K Q _{0} ( - q )}{l} \times 2 + \frac{K ( - q ) ( - q )}{l 2} = 0$
$\frac{K q}{l} [- 2 Q_{0} + \frac{q}{2}] = 0$
$∴$ $Q_{0} = \frac{q}{2 2} = \frac{q}{8}$

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​

8​q​

2​q

q

U=lKQ0​(−q)​×2+l2​K(−q)(−q)​=0 lKq​[−2Q0​+2​q​]=0 ∴ Q0​=22​q​=8​q​

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

$\frac{q}{4}$

$\frac{q}{8}$

$2 q$

$q$

$U = \frac{K Q _{0} ( - q )}{l} \times 2 + \frac{K ( - q ) ( - q )}{l 2} = 0$
$\frac{K q}{l} [- 2 Q_{0} + \frac{q}{2}] = 0$
$∴$ $Q_{0} = \frac{q}{2 2} = \frac{q}{8}$