Question

A small metal sphere, carrying a net charge of $q_1=-2\mu C$, is held in a stationery position by insulating supports. A second small metal sphere, with a net charge of $q_2=-8\mu C$ and mass 1.50g is projected towards $q_1$. When the two spheres are $0.800\text{m}$ apart, $q_2$ is moving towards $q_1$ with speed $20m{s}^{-1}$ as shown in the figure. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.
The speed of $q_2$ when the spheres are 0.400 m apart is

• A. $2\sqrt{10}m{s}^{-1}$
• B. $2\sqrt{6}m{s}^{-1}$
• C. $4\sqrt{10}m{s}^{-1}$
• D. $4\sqrt{6}m{s}^{-1}$

Answer 1

The correct option is C $4\sqrt{10}m{s}^{-1}$

Let $i$ be the point when the charge $q_2$ is $0.8m$ apart and $f$ be the point when it is $0.4m$ apart.

As both the charges are of same nature so there will be electrostatic force of replusion on charge $q_2$ due to $q_1$ which will decrease the velocity of charge $q_2$.

We know that electrostatic force is a conservative force so the total energy will remain conserved.
$⟹E_i=E_f$
$P{E}_i+K{E}_i=P{E}_i+K{E}_f$
Velocity of charge $q_2$ at point $i$ = $20m{s}^{-1}$
$\therefore$ Kinetic energy of $q_2$ at point $i$ = $\frac{1}{2}m(20{)}^2$ $J$
Potential energy of charge $q_2$ at point $i$ will be $\frac{k{q}_1{q}_2}{0.8}$
similarly, potential energy at point $f$ will be $\frac{k{q}_1{q}_2}{0.4}$
Let velocity of $q_2$ at point $f$ be $v_f$
$⟹\frac{k{q}_1{q}_2}{0.8}+\frac{1}{2}m(20{)}^2=\frac{k{q}_1{q}_2}{0.4}+\frac{1}{2}m{v}_f^2$
$⟹\frac{9\times {10}^9\times 2\times {10}^{-6}\times 8\times {10}^{-6}}{0.8}+\frac{1}{2}\left(0.0015\right)(20{)}^2=\frac{9\times {10}^9\times 2\times {10}^{-6}\times 8\times {10}^{-6}}{0.4}+\frac{1}{2}m{v}_f^2$
$⟹0.48J=0.36J+\frac{1}{2}m{v}_f^2$
$⟹v_f=\sqrt{\frac{\left(0.12\right)\times 2}{0.0015}}$
$\therefore v_f=4\sqrt{10}m{s}^{-1}$