Two point charges +q_{1} and +q_{2} are place | vedclass

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(C) For a charge $q_{3}$ to be in equilibrium,the net electrostatic force acting on it must be zero.Let the distance of $q_{3}$ from $q_{1}$ be $x$. T

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possible irrespective of the sign of $q_{3}$.

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(C) For a charge $q_{3}$ to be in equilibrium,the net electrostatic force acting on it must be zero.Let the distance of $q_{3}$ from $q_{1}$ be $x$. The force due to $q_{1}$ is $F_{1} = \frac{k q_{1} q_{3}}{x^{2}}$ and the force due to $q_{2}$ is $F_{2} = \frac{k q_{2} q_{3}}{(d-x)^{2}}$.For equilibrium,$F_{1} = F_{2}$ (in magnitude and opposite in direction).$\frac{k q_{1} q_{3}}{x^{2}} = \frac{k q_{2} q_{3}}{(d-x)^{2}}$$\frac{q_{1}}{x^{2}} = \frac{q_{2}}{(d-x)^{2}}$This equation shows that the position $x$ depends only on the magnitudes of $q_{1}$ and $q_{2}$ and the distance $d$. The charge $q_{3}$ cancels out from both sides of the equation.Therefore,the equilibrium position is independent of the sign and magnitude of the third charge $q_{3}$.Thus,it is possible irrespective of the sign of $q_{3}$.

Two point charges $+q_{1}$ and $+q_{2}$ are placed a finite distance '$d$' apart. It is desired to put a third charge $q_{3}$ in between these two charges so that $q_{3}$ is in equilibrium. This is

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