Question

Two fixed charges $A$ and $B$ of $5\mu C$ each are separated by a distance of $6m$. $C$ is the mid point of the line joining $A$ and $B$. A charge ${}^{\prime }{Q}^{\prime }$ of $-5\mu C$ is shot perpendicular to the line joining $A$ and $B$ through $C$ with a kinetic energy of $0.06J$. The charge ${}^{\prime }{Q}^{\prime }$ comes to rest at a point $D$. The distance $CD$ is

• A. $3m$
• B. $\sqrt{3}m$
• C. $3\sqrt{3}m$
• D. $4m$

Answer 1

Answer: (D)

Initial potential energy of the system is given by:
$U=\frac{K{Q}_1{Q}_2}{r_{12}}+\frac{K{Q}_2{Q}_3}{r_{23}}+\frac{K{Q}_3{Q}_1}{r_{13}}$
At $D$, charge of $-5\mu C$ comes to rest. So, total energy equals potential energy at this point.
Applying conservation of energy,
$\begin{array}{rl}& E_i=E_f\\ \Rightarrow -0.0525=9\times 10^9\times & \\ \left(\frac{5\times 10^{-6}\times 5\times 10^{-6}}{6}+\frac{5\times 10^{-6}\times -5\times 10^{-6}}{AD}+\frac{5\times 10^{-6}\times -5\times 10^{-6}}{BD}\right)& \end{array}$
Using symmetry, $AD=BD$
Solving, we get
$AD=5m$ Substituting values,
$\begin{array}{l}U=9\times 10^9\times \left(\frac{5\times 10^{-6}\times 5\times 10^{-6}}{6}+\frac{5\times 10^{-6}\times -5\times 10^{-6}}{3}+\frac{5\times 10^{-6}\times -5\times 10^{-6}}{3}\right)\\ U=-0.1125J\end{array}$
Total initial energy of the system is equal to the sum of potential energy and kinetic energy.
$E=U+K=-0.1125+0.06=-0.0525J$