Three charges q, Q and +4q are placed in a st | vedclass

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(NONE) Let the positions of the charges be $x_1 = 0$, $x_2 = d/3$, and $x_3 = 2d/3$.The charge at $x_1$ is $q$, at $x_2$ is $Q$, and at $x_3$ is $+4q$

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

Vedclass · Answer

(NONE) Let the positions of the charges be $x_1 = 0$, $x_2 = d/3$, and $x_3 = 2d/3$.The charge at $x_1$ is $q$, at $x_2$ is $Q$, and at $x_3$ is $+4q$.For the net force on the charge $q$ at $x_1 = 0$ to be zero, the force exerted by $Q$ and the force exerted by $+4q$ must be equal in magnitude and opposite in direction.The force exerted by $Q$ on $q$ is $F_1 = \frac{k q Q}{(d/3)^2} = \frac{9 k q Q}{d^2}$.The force exerted by $+4q$ on $q$ is $F_2 = \frac{k q (4q)}{(2d/3)^2} = \frac{k q (4q)}{4d^2/9} = \frac{9 k q^2}{d^2}$.For the net force to be zero, $F_1 + F_2 = 0$, which implies $F_1 = -F_2$.$\frac{9 k q Q}{d^2} = -\frac{9 k q^2}{d^2}$.Canceling common terms $9kq/d^2$ from both sides, we get $Q = -q$.

Three charges $q, Q$ and $+4q$ are placed in a straight line of length $d$ at positions $0, d/3$ and $2d/3$ respectively. In order to make the net force on $q$ be zero, the value of $Q$ should be

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