Question

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

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

Answer 1

Answer: (D)

Initial potential energy of the system is given by:
$ U =\frac{ KQ _1 Q _2}{ r _{12}}+\frac{ KQ _2 Q _3}{ r _{23}}+\frac{ KQ _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{aligned} & 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{aligned} $
Using symmetry, $AD = BD$
Solving, we get
$ AD =5 m $ 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=-0.1125 J \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=-0.0525 J $