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 $C O D$, the perpendicular bisector of the line $A B$. The moving charge, when reaches the point $C$ at a distance of $4\, m$ from $O$, has a kinetic energy of $4 \,J$. Calculate the distance $x$ (in $m$ ) of the farthest point $D$ at which the negative charge will reach before returning towards $C$.

Answer 1

Figure shows the situation described in question.

From figure, we have
$ A C=\sqrt{(A O)^{2}+(O C)^{2}} $
$\Rightarrow A C=\sqrt{(3)^{2}+(4)^{2}}=5\,m $
Similarly, we have $B C=5\, m$
Potential energy of charge $-q $ at point $ C $ is given as
$U_{C}=-\frac{2 K q^{2}}{(A C)}=-2 \times\left(9 \times 10^{9}\right) \times \frac{q^{2}}{A C}$
$\Rightarrow U_{C}=-2 \times\left(9 \times 10^{9}\right) \frac{\left(5 \times 10^{-5}\right)^{2}}{5}$
$\Rightarrow U_{C}=-9\, J$
Kinetic energy of charge at $C$ is given as $K_{c}=4 \,J$
Total energy of charge at $C$ is given as,
$T E=P E+K E$
$\Rightarrow T E=-9+4=-5 \,J$
If we use $A D=B D=r$ then potential energy of charge at point $D$ is given as
$U_{D} =-\frac{2 K q^{2}}{r} $
$\Rightarrow U_{D} =\frac{-2 \times\left(9 \times 10^{9}\right)\left(5 \times 10^{-5}\right)^{2}}{r} $
$\Rightarrow U_{D} =\frac{-45}{r} J$
As kinetic energy of charge at point $D$ is zero, by conservation of energy at point $C$ and $D$, we use
$\frac{-45}{r}=-5 \Rightarrow r=9 \,m$
From figure, we use $x=O D=\sqrt{(A D)^{2}-(A O)^{2}}$
$\Rightarrow x=\sqrt{(9)^{2}-(3)^{2}}$
$=6 \sqrt{2} m =8.49 \,m$