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 $6m$. An equal and opposite charge moves towards them along the line $COD$, the perpendicular bisector of the line $AB$. The moving charge, when reaches the point $C$ at a distance of $4m$ from $O$, has a kinetic energy of $4J$. Calculate the distance $x$ (in $m$ ) of the farthest point $D$ at which the negative charge will reach before returning towards $C$.

Answer 1

Answer: 8.49

Figure shows the situation described in question.

From figure, we have
$AC=\sqrt{(AO{)}^2+(OC{)}^2}$
$\Rightarrow AC=\sqrt{(3{)}^2+(4{)}^2}=5m$
Similarly, we have $BC=5m$
Potential energy of charge $-q$ at point $C$ is given as
$U_C=-\frac{2K{q}^2}{(AC)}=-2\times \left(9\times 10^9\right)\times \frac{q^2}{AC}$
$\Rightarrow U_C=-2\times \left(9\times 10^9\right)\frac{{\left(5\times 10^{-5}\right)}^2}{5}$
$\Rightarrow U_C=-9J$
Kinetic energy of charge at $C$ is given as $K_c=4J$
Total energy of charge at $C$ is given as,
$TE=PE+KE$
$\Rightarrow TE=-9+4=-5J$
If we use $AD=BD=r$ then potential energy of charge at point $D$ is given as
$U_D=-\frac{2K{q}^2}{r}$
$\Rightarrow U_D=\frac{-2\times \left(9\times 10^9\right){\left(5\times 10^{-5}\right)}^2}{r}$
$\Rightarrow U_D=\frac{-45}{r}J$
As kinetic energy of charge at point $D$ is zero, by conservation of energy at point $C$ and $D$, we use
$\frac{-45}{r}=-5\Rightarrow r=9m$
From figure, we use $x=OD=\sqrt{(AD{)}^2-(AO{)}^2}$
$\Rightarrow x=\sqrt{(9{)}^2-(3{)}^2}$
$=6\sqrt{2}m=8.49m$