Two charges +q and -3q are placed on the x-ax | vedclass

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(A) Let the third charge $2q$ be placed at a distance $x$ to the left of charge $+q$. For the net force on $2q$ to be zero,the magnitude of the electr

Vedclass · Accepted Answer

$\frac{d}{2}(1 + \sqrt{3})$ to the left of $q$

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(A) Let the third charge $2q$ be placed at a distance $x$ to the left of charge $+q$. For the net force on $2q$ to be zero,the magnitude of the electrostatic force due to $+q$ must equal the magnitude of the electrostatic force due to $-3q$. Using Coulomb's law: $\frac{k(q)(2q)}{x^2} = \frac{k(3q)(2q)}{(x+d)^2}$. Simplifying the equation: $\frac{1}{x^2} = \frac{3}{(x+d)^2}$. Taking the square root on both sides: $\frac{1}{x} = \frac{\sqrt{3}}{x+d}$. $x+d = x\sqrt{3} \Rightarrow d = x(\sqrt{3}-1)$. $x = \frac{d}{\sqrt{3}-1}$. Rationalizing the denominator: $x = \frac{d(\sqrt{3}+1)}{3-1} = \frac{d}{2}(1+\sqrt{3})$. Thus,the charge should be placed at a distance $\frac{d}{2}(1+\sqrt{3})$ to the left of $q$.

Two charges $+q$ and $-3q$ are placed on the $x$-axis separated by a distance $d$. ($-3q$ is to the right of $q$). Where should a third charge $2q$ be placed such that it will not experience any net force?

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