Two free point charges +q and +4q are a dista | vedclass

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(D) Let the third charge $Q$ be located at a distance $x$ from the $+q$ charge. For the third charge to be in equilibrium,the force on it due to the $

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$-\frac{4}{9} q$

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(D) Let the third charge $Q$ be located at a distance $x$ from the $+q$ charge. For the third charge to be in equilibrium,the force on it due to the $+q$ charge and $+4q$ charge must be equal in magnitude and opposite in direction.$\frac{kqQ}{x^2} = \frac{kQ(4q)}{(R-x)^2}$$\frac{1}{x^2} = \frac{4}{(R-x)^2}$Taking the square root on both sides: $\frac{1}{x} = \frac{2}{R-x}$$R-x = 2x \implies x = \frac{R}{3}$For the entire system to be in equilibrium,the net force on the $+q$ charge must also be zero:$\frac{kq(4q)}{R^2} + \frac{kqQ}{x^2} = 0$$\frac{4kq^2}{R^2} + \frac{kqQ}{(R/3)^2} = 0$$\frac{4q}{R^2} + \frac{9Q}{R^2} = 0$$4q + 9Q = 0 \implies Q = -\frac{4}{9}q$

Two free point charges $+q$ and $+4q$ are a distance $R$ apart. $A$ third charge is so placed that the entire system is in equilibrium. Then the third charge is :-

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