You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.30 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.10 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.15 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $f$ = $0$ $∵ v_{f} = 0$
$⟹ \frac{k q_{1} q_{2}}{0.8} + \frac{1}{2} (0.0015) (20)^{2} = \frac{k q_{1} q_{2}}{r} + 0$
$⟹ 0.48 J = \frac{k q_{1} q_{2}}{r}$
$⟹ r = \frac{9 \times 10^{9} \times 2 \times 10^{- 6} \times 8 \times 10^{- 6}}{0.48}$
$∴ r = 0.3 m$

Suggest Corrections

Similar questions

Two uniformly charged spheres with radii $r_{1}$ & $r_{2}$ and charges $Q_{1}$ & $Q_{2}$ are separated by distance d (from the periphery of the two spheres). What is the magnitude of force on $Q_{2}$ due to $Q_{1}$ ?

Q. Two identical small conducting spheres carry charges of

Q_{1}

and

Q_{2}

with

Q_{1} >> Q_{2}

. The spheres are d distance apart. The force they exert on each other is

F_{1}

. The spheres are made to touch each other and then separated by a distance d. The force they exert on each other now is

F_{2}

. Then

F_{1} / F_{2}

is :-

Q. Two positive charges

q_{1}

and

q_{2}

are moving with velocities

v_{1}

and

v_{2}

when they are at points

A

and

B

, respectively, as shown in figure. The magnetic force experienced by charge

q_{1}

due to the other charge

q_{2}

is :

Q. For the given charge configuration consisting of charges

+ q_{1}

and

+ q_{2}

such that

q_{1} > q_{2}

, the relation between flux through two imaginary spheres

S_{1}

and

S_{2}

as shown in the figure will be

Q. A small sphere of radius

r_{1}

and charge

q_{1}

is enclosed by a spherical shell of radius

r_{2}

and

q_{2}

. charged sphere and shell are connnected by a wire. if

q_{1}

is positive then

Join BYJU'S Learning Program