Question

Two fixed, equal, positive charges, each of magnitude $5\times {10}^{-5}C$ are located at points A and B separated by a distance of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge when it reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 J. If the distance of the farthest point D which the negative charge will reach before returning towards C is $x$, then find the value of $x^2/12$.

Answer 1

Equating the energy of $(-q)$ at $C$ with the energy at $D$ i.e, $K_C+U_C=K_D+U_D$

Here, $K_C=4J$

$U_C=2\left[\frac{1}{4\pi {ϵ}_0}\frac{\left(q\right)(-q)}{AC}\right]$

$=\frac{-2\times 9\times {10}^9\times ({5\times {10}^{-5})}^2}{5}$

=$-9J$

$K_D=0$ and $U_D=2\left[\frac{1}{4\pi {ϵ}_0}\frac{\left(q\right)(-q)}{AC}\right]$

$=\frac{-2\times 9\times {10}^9\times ({5\times {10}^{-5})}^2}{AD}$

$=\frac{-45}{AD}$

$⟹4-9=0-\frac{45}{AD}$

$\therefore AD=9m$

$\therefore OD=\sqrt{A{D}^2-O{A}^2}=\sqrt{9^2-3^2}=8.48m=x$

$\therefore \frac{x^2}{12}=5.99\approx 6m^2$