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Potential Energy of Two Point Charges

A small metal...

Question

A small metal sphere, carrying a net charge of $q_{1} = - 2 μ C$ , is held in a stationery position by insulating supports. A second small metal sphere, with a net charge of $q_{2} = - 8 μ C$ and mass $1.50 g$ is projected towards $q_{1}$ . When the two spheres are $0.800 m$ apart, $q_{2}$ is moving towards $q_{1}$ with speed $20 m s^{- 1}$ as shown in the figure. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.
How close does $q_{2}$ get to $q_{1}$ ?

0.20 m

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0.30 m

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0.10 m

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0.15 m

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Solution

The correct option is B $0.30 m$
Let $i$ be the point when the charge $q_{2}$ is $0.8 m$ .
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 $q_{2}$ .
Ultimately, charge $q_{2}$ will reach point a $f$ where it is at minimum distance with $q_{1}$ and its velocity becomes $0$ at this point.

Let this minimum distance between $q_{1}$ and $q_{2}$ be $r$ .
By total energy conservation
$⟹ E_{i} = E_{f}$
$P E_{i} + K E_{i} = P E_{i} + K E_{f}$
Velocity of charge $q_{2}$ at point $i$ = $20 m s^{- 1}$
$∴$ 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 $\frac{k q_{1} q_{2}}{0.8}$
Potential energy of charge $q_{2}$ at point $f$ will be $\frac{k q_{1} q_{2}}{r}$
Kinetic energy of $q_{2}$ at point $f$ = $0$ $∵ v_{f} = 0$
$⟹ \frac{k q_{1} q_{2}}{0.8} + \frac{1}{2} (0.0015) (20)^{2} = \frac{k q_{1} q_{2}}{r} + 0$
$⟹ 0.48 J = \frac{k q_{1} q_{2}}{r}$
$⟹ r = \frac{9 \times 10^{9} \times 2 \times 10^{- 6} \times 8 \times 10^{- 6}}{0.48}$
$∴ r = 0.3 m$

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