Three charges +Q, q, + Q are placed respectively, at distance, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x = 0, Ls zero, then value of q is :

Question

Three charges +Q, q, + Q are placed respectively, at distance, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x = 0, Ls zero, then value of q is : (A) +Q/2 (B) -Q/2 (C) -Q/4 (D) +Q/4

Tardigrade Admin · Accepted Answer

For equilibrium, vecFa + vecFB = 0 vecFa = - vecFB (kQQ/d2) = - (kQq/(d/2)2) ⇒ q = - (Q/4)

Q. Three charges $+ Q, q, + Q$ are placed respectively, at distance, $d /2$ and $d$ from the origin, on the $x$ -axis. If the net force experienced by $+ Q$ , placed at $x = 0$ , $L s$ zero, then value of $q$ is :

+Q/2

-Q/2

-Q/4

+Q/4

For equilibrium,
$F_{a} + F_{B} = 0$
$F_{a} = - F_{B}$
$\frac{k QQ}{d ^{2}} = - \frac{k Qq}{( d /2 ) ^{2}}$
$\Rightarrow q = - \frac{Q}{4}$

Q. Three charges +Q,q,+Q are placed respectively, at distance, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x=0, Ls zero, then value of q is :