The figure shows two equal, positive charges, each of magnitude $50\mathrm{\mu C}$, fixed at points (3, 0) m and (–3, 0)m respectively. A charge $-50\mathrm{\mu C}$, moving along negative y–axis has a kinetic energy of 4J at the instant it crosses point (0,4)m. Determine the position of this charge where the direction of its motion reverses for the first time after crossing this point (neglect gravity).

The charge $-50\mathrm{\mu C}$ will move in straight line along y-axis as it does not experience any force in x-direction.
Let B be the location where the charge comes to rest momentarily and then return. Total energy of the system remain constant.
$\therefore$ KE + PE
$\begin{array}{l}=4+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(50\times {10}^{-6}\right)\left(-50\times {10}^{-6}\right)}{5}\times 2\\ =0+\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(50\times {10}^{-6}\right)\left(-50\times {10}^{-6}\right)}{\sqrt{3^2+{\mathrm{y}}^2}}\times 2\\ \frac{1}{4{\mathrm{\pi \epsilon }}_0}=9\times {10}^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2}\end{array}$
 

$\therefore$ Solving for y
we get y = $6\sqrt{2}\mathrm{m}$. (since body is going down negative value is chosen)
$\therefore$ The location is $(0,-6\sqrt{2}\mathrm{m})$.