Two point charges +e and +4e are kept at a di | vedclass

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Question

(A) For the third charge $+q$ to be in equilibrium, the electrostatic force exerted by $+e$ on $+q$ must be equal in magnitude to the force exerted by

Vedclass · Accepted Answer

$\frac{d}{3}$ units

Vedclass · Answer

(A) For the third charge $+q$ to be in equilibrium, the electrostatic force exerted by $+e$ on $+q$ must be equal in magnitude to the force exerted by $+4e$ on $+q$.Let the distance between $+e$ and $+q$ be $x$. Then the distance between $+4e$ and $+q$ is $(d - x)$.Using Coulomb's Law, the force $F_1$ due to $+e$ is $F_1 = \frac{k \cdot e \cdot q}{x^2}$.The force $F_2$ due to $+4e$ is $F_2 = \frac{k \cdot 4e \cdot q}{(d - x)^2}$.Setting $F_1 = F_2$, we get: $\frac{k \cdot e \cdot q}{x^2} = \frac{k \cdot 4e \cdot q}{(d - x)^2}$.Canceling common terms $k, e, q$, we have: $\frac{1}{x^2} = \frac{4}{(d - x)^2}$.Taking the square root of both sides: $\frac{1}{x} = \frac{2}{d - x}$.Cross-multiplying gives: $d - x = 2x$.Therefore, $3x = d$, which implies $x = \frac{d}{3}$.

Two point charges $+e$ and $+4e$ are kept at a distance $d$ units apart. $A$ third point charge $+q$ is placed between the two charges at a distance $x$ units from charge $+e$ so as to be in equilibrium. The value of $x$ is:

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