Question

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

Answer 1

Just before reversing the direction of motion, speed of $-50\mu C$ point charge must be zero. [say at $(0,-y)$]
Let us apply conservation of mechanical energy of system.
$4+\frac{k50\times {10}^{-6}}{5}\times (-50\times {10}^{-6})\times 2=0+k\frac{50\times {10}^{-6}}{\sqrt{y^2+9}}\times (-50\times {10}^{-6})\times 2$
On solving
$4-9=\frac{-45}{\sqrt{y^2+9}}$
$\Rightarrow \sqrt{y^2+g}=g$
$\Rightarrow y^2=72$
$\Rightarrow y=6\sqrt{2}=3\sqrt{8}$