Question

Two equal charges A and B each of $1/3\times {10}^{-6}C$ are placed 200 cm apart in air. A particle carrying a charge of $-1/3\times {10}^{-6}C$ is projected along the perpendicular bisector from the point O midway between A and B with a kinetic energy of ${10}^{-3}J$. Before the particle starts to return it will cover a distance

• A. $1m$
• B. $\sqrt{2}$ m
• C. $\sqrt{3}$ m
• D. $1/\sqrt{3}$ m

Answer 1

Answer: (C)

$\sqrt{3}$ m

Initial potential energy: $\frac{2k{q}_1{q}_2}{r}=(2\times 9\times {10}^9\times \frac{1}{3}\times {10}^{-6}\times \frac{-1}{3}\times {10}^{-6})/\left(1\right)=-2\times {10}^{-3}$

Initial kinetic energy is 10^(-3) J.

Let the body is at r distance from both the charges (will be at equal distance from both as both charges are same)

Final Kinetic energy is zero (as body stopped).

By equating Initial Total Energy & Final Total Energy:

$-2\times {10}^{-3}+{10}^{-3}=\frac{2k{q}_1{q}_2}{r}=(2\times 9\times {10}^9\times \frac{1}{3}\times {10}^{-6}\times \frac{-1}{3}\times {10}^{-6})/\left(r\right)-1\times {10}^{-3}=-2\times {10}^{-3}/\left(r\right)\therefore r=2$

Therefore distance traveled:

$\sqrt{(r_f^2-r_i^2)}=\sqrt{4-1}=\sqrt{3}$