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 '$Q$' 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 '$Q$' 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

The correct option is D $4m$

$Step1:Initialenergyofthesystem$

Initial potential energy of the system is given by:

$U=\frac{K{Q}_1{Q}_2}{r_{12}}+\frac{K{Q}_2{Q}_3}{r_{23}}+\frac{K{Q}_3{Q}_1}{r_{13}}$

Substituting values,

$U_i=9\times {10}^9\times (\frac{5\times {10}^{-6}C\times 5\times {10}^{-6}C}{6m}+\frac{5\times {10}^{-6}C\times -5\times {10}^{-6}C}{3m}+\frac{5\times {10}^{-6}C\times -5\times {10}^{-6}C}{3m})$

⇒$U_i=-0.1125J$

Total initial energy of the system is equal to the sum of potential energy and kinetic energy.

$E_i=U_i+K_i=-0.1125+0.06=-0.0525J$

$Step2:Finalenergyofthesystem$

Final potential energy of system is given by:

$U=\frac{K{Q}_1{Q}_2}{r_{12}}+\frac{K{Q}_2{Q}_3}{r_{23}}+\frac{K{Q}_3{Q}_1}{r_{13}}$

$U_f=9\times {10}^9\times (\frac{5\times {10}^{-6}C\times 5\times {10}^{-6}C}{6m}+\frac{5\times {10}^{-6}C\times -5\times {10}^{-6}C}{AD}+\frac{5\times {10}^{-6}C\times -5\times {10}^{-6}C}{BD})$

Using symmetry, $AD=BD$

⇒$U_f=9\times {10}^{-3}(4.16-\frac{50}{AD})$

Total final energy of the system will be equal to the sum of potential energy and kinetic energy.

$E_f=U_f+K_f$

Finally charge is at rest

So, $K_f=0$

⇒$E_f=9\times {10}^{-3}(4.16-\frac{50}{AD})$

$Step3:EnergyConservation$

Electrostatic force is a conservative force so mechanical energy will be conserved

Suppose at D, charge of $-5\mu C$ comes to rest

Applying conservation of energy,

$E_i=E_f$

$\Rightarrow -0.0525=9\times {10}^{-3}(4.16-\frac{50}{AD})$

$\Rightarrow AD=5m$

Using Pythagoras theorem,

${AD}^2={AC}^2+{CD}^2$

$\Rightarrow 5^2=3^2+{CD}^2$

$\Rightarrow CD=4m$

Hence, option D is correct.