Two point charges +4 e and + e are kept at distance x apart. At what distance a charge q must be placed from charge + e, so that q is in equilibrium :-

Question

Two point charges +4 e and + e are kept at distance x apart. At what distance a charge q must be placed from charge + e, so that q is in equilibrium :- (A) ( x /2) (B) (2 x/3) (C) ( x /3) (D) (x/6)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/29d60d096d212a61cc3e36b2a4e6d854-.png" /> If q is in equilibrium at position y from +e then (+ e /4 π ∈0 y 2)=(+4 e /4 π ∈0( x - y )2) ⇒ x-y=2 y ⇒ y=(x/3)

Q. Two point charges $+ 4 e$ and $+ e$ are kept at distance $x$ apart. At what distance a charge $q$ must be placed from charge $+ e$ , so that $q$ is in equilibrium :-

$\frac{x}{2}$

$\frac{2 x}{3}$

$\frac{x}{3}$

$\frac{x}{6}$

If $q$ is in equilibrium at position y from $+ e$ then
$\frac{+ e}{4 π \in _{0} y ^{2}} = \frac{+ 4 e}{4 π \in _{0} ( x - y ) ^{2}}$
$\Rightarrow x - y = 2 y$
$\Rightarrow y = \frac{x}{3}$

Q. Two point charges +4e and +e are kept at distance x apart. At what distance a charge q must be placed from charge +e, so that q is in equilibrium :-

2x​

32x​

3x​

6x​