Question

Two fixed, equal, positive charges, each of magnitude $5\times {10}^{-5}C$ are located at point $A$ and $B$ separated by a distance $6m$. 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 $4m$ from $O$, has a kinetic energy of $4$ joules. Calculate the distance of the farthest point $D$ which the negative charge will reach before returning towards $C$.

Answer 1

Equating the energy og $(-q)$ at $C$ & $D$

$k_c+v_c=k_D+v_D$

Here $k_c=4J$

$v_c=2\left[\frac{\frac{1}{4\pi \epsilon }q(-q)}{A_c}\right]$

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

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

$=-9J$

$k_D=0$

$v_D=2\left[\frac{1}{4\pi {\epsilon }_0}\frac{q(-q)}{AD}\right]$

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

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

$4-9=0-\frac{45}{AD}$

$AD=9m$

$OD=\sqrt{A{D}^2-D{A}^2}$

$=\sqrt{(9{)}^2-(3{)}^2}$

$=\sqrt{81-9}=8.48m$