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

Charging by Induction

A positively ...

Question

A positively charged disc is placed on a horizontal plane. A charged particle is released from a certain height on its axis. The particle just reaches the centre of the disc then

particle has a negative charge on it.

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

total potential energy (gravitational + electrostatic) first increases and then decreases.

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

total potential energy (gravitational + electrostatic) first decreases and then increases.

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

total potential energy (gravitational + electrostatic) decreases continuously.

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

Open in App

Solution

The correct option is C total potential energy (gravitational + electrostatic) first decreases and then increases.
The particle is positively charged because it is just reaching the centre of the disc $i . e .$ kinetic energy is zero which is only possible when there is repulsive force on the particle.

Also, as the particle descends, its total potential energy (gravitational + electrostatic) first decreases and provide kinetic energy to the particle. In this phase, gravitational force on the particle is more than the electrostatic repulsive force.

As the particle approaches the disc, electrostatic repulsive force become stronger and stronger and starts retarding the particle after certain time. In this phase, kinetic energy decreases and total potential energy increases.

Hence, total potential energy (gravitational + electrostatic) first decreases and then increases.

Suggest Corrections

Similar questions

Q. A non conducting disc of radius

a

and uniform positive surface charge density

σ

is placed on the ground with its axis vertical. A particle of mass

m

and positive charge

q

is dropped along axis of disc from a height

H

with zero initial velocity. The particle has

\frac{q}{m} = \frac{4 ε_{0} g}{σ}

. The value of

H

if the particle just reaches the disc is

Q. A non conducting disc of radius

a

and uniform positive surface charge density

σ

is placed on the ground with its axis vertical. A particle of mass

m

and charge

q

is dropped, along the axis of the disc from a height

H

with zero initial velocity. It was observed that its velocity becomes zero just when it reaches the disc. The value of

\frac{3 H}{a}

is (Given

\frac{q}{m} = \frac{4 ϵ_{o} g}{σ}

)

Q. A non conducting disc of radius

a

and uniform positive surface charge density

σ

is placed on the ground with its axis vertical. A particle of mass

m

and charge

q

is dropped, along the axis of the disc from a height

H

with zero initial velocity. It was observed that its velocity becomes zero just when it reaches the disc. The value of

\frac{3 H}{a}

is (Given

\frac{q}{m} = \frac{4 ϵ_{o} g}{σ}

)

Q. A thin disc of radius

R = 50

cm with a circular hole (of radius

r = 0.1 R

) at its centre is charged to a uniform positive surface charge density

10 μ C m^{- 2}

and is placed in y-z plane with the origin at the centre of the disc as shown in the figure. A particle of mass

1 g

with charge

- 10 m C

which is free to move only along the x-axis, is placed at a distance

x = 0.01 R

and released at

t = 0 s

. Then:

Q. A circular disc of radius

^{'} R^{'}

and mass

^{'} m^{'}

is hinged at the centre about horizontal axis. A particle of same mass is attached at the top of the disc. Now the system is released. Calculate the angular velocity of the disc when the particle reaches bottom.

Join BYJU'S Learning Program