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(C) The net force on charge $Q$ at a distance $x$ from the origin on the $X$-axis is given by the sum of the forces from the two charges $q$. The dist

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Move to infinity

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(C) The net force on charge $Q$ at a distance $x$ from the origin on the $X$-axis is given by the sum of the forces from the two charges $q$. The distance of $Q$ from each charge $q$ is $r = \sqrt{x^2 + a^2}$.The force from each charge is $F = \frac{1}{4\pi\epsilon_0} \frac{qQ}{x^2 + a^2}$.The component of this force along the $X$-axis is $F_x = F \cos	heta = F \frac{x}{r} = \frac{1}{4\pi\epsilon_0} \frac{qQx}{(x^2 + a^2)^{3/2}}$.Since there are two such charges,the net force is $F_{net} = 2 F_x = \frac{1}{2\pi\epsilon_0} \frac{qQx}{(x^2 + a^2)^{3/2}}$.This force is directed away from the origin (repulsive). Since the charge $Q$ is released at $(2a, 0)$,it will be pushed further away from the origin along the $X$-axis towards infinity. Therefore,it will move to infinity.

Two equal positive charges $q$ are fixed at the points $(0, a)$ and $(0, -a)$ on the $Y$-axis. $A$ positive charge $Q$ is released from rest at the point $(2a, 0)$ on the $X$-axis. The charge $Q$ will:

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