Two free positive charges 4q and q are at a d | vedclass

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(A) For the system to be in equilibrium,the net force on each charge must be zero.Let the charge $Q$ be placed at a distance $x$ from charge $q$ and $

Vedclass · Accepted Answer

$Q = \frac{4}{9} q$ (negative) at $\frac{l}{3}$ distance

Vedclass · Answer

(A) For the system to be in equilibrium,the net force on each charge must be zero.Let the charge $Q$ be placed at a distance $x$ from charge $q$ and $(l-x)$ from charge $4q$.For charge $q$ to be in equilibrium,the forces from $4q$ and $Q$ must balance:$\frac{k(4q)(q)}{l^2} = \frac{k(Q)(q)}{x^2} \implies \frac{4}{l^2} = \frac{Q/q}{x^2} \implies \frac{Q}{q} = 4\frac{x^2}{l^2} \quad (1)$For charge $4q$ to be in equilibrium,the forces from $q$ and $Q$ must balance:$\frac{k(4q)(q)}{l^2} = \frac{k(4q)(Q)}{(l-x)^2} \implies \frac{q}{l^2} = \frac{Q}{(l-x)^2} \implies \frac{Q}{q} = \frac{(l-x)^2}{l^2} \quad (2)$Equating $(1)$ and $(2)$:$4\frac{x^2}{l^2} = \frac{(l-x)^2}{l^2} \implies 4x^2 = (l-x)^2 \implies 2x = l-x \implies 3x = l \implies x = \frac{l}{3}$.Substituting $x = l/3$ into $(1)$:$\frac{Q}{q} = 4 \frac{(l/3)^2}{l^2} = 4 \cdot \frac{1}{9} = \frac{4}{9}$.Since the charges $4q$ and $q$ are positive,$Q$ must be negative to provide an attractive force to balance the repulsion between them. Thus,$Q = -\frac{4}{9}q$.

Two free positive charges $4q$ and $q$ are at a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed from charge $q$?

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