Two equal charges q are placed at x = -a and | vedclass

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(A) The net force on the charge $q_0$ at a distance $y$ from the origin along the $y$-axis is due to the two charges $q$ at $(-a, 0)$ and $(a, 0)$.By

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$y$

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(A) The net force on the charge $q_0$ at a distance $y$ from the origin along the $y$-axis is due to the two charges $q$ at $(-a, 0)$ and $(a, 0)$.By symmetry,the horizontal components of the forces cancel out,and the vertical components add up.The force from each charge $q$ on $q_0$ is $F = \frac{k q q_0}{r^2}$,where $r^2 = a^2 + y^2$.The net force is $F_{net} = 2F \sin	heta$,where $\sin	heta = \frac{y}{r} = \frac{y}{\sqrt{a^2 + y^2}}$.Substituting the values,we get $F_{net} = 2 \left[ \frac{k q q_0}{a^2 + y^2} ight] \cdot \frac{y}{(a^2 + y^2)^{1/2}} = \frac{2 k q q_0 y}{(a^2 + y^2)^{3/2}}$.For a small displacement $(y Thus,$F_{net} \approx \frac{2 k q q_0 y}{a^3}$.Since $k, q, q_0,$ and $a$ are constants,the net force is proportional to $y$.

Two equal charges $q$ are placed at $x = -a$ and $x = a$ on the $x$-axis. $A$ particle of mass $m$ and charge $q_0 = q/2$ is placed at the origin. If the charge $q_0$ is given a small displacement $(y << a)$ along the $y$-axis,the net force acting on the particle is proportional to .......

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