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

Byju's Answer

Standard XII

Physics

Analysis of Force Equation

An oil drop o...

Question

An oil drop of mass m and charge $+ q$ is balanced in vacuum by a uniform electric field of intensity $E$ The direction of this field should be

vertically up.

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

vertically down.

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

horizontal.

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

inclined at

45^{o}

to the horizontal.

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

Open in App

Solution

The correct option is A vertically up.
$m g = q E$
$E$ should be upward to balance downward $m g$ .

Suggest Corrections

Similar questions

Q. An oil drop of radius r carrying a charge 'q' remains stationary in the presence of electric field of intensity E. If the density of oil is

ρ

, then

Q. A particle of mass m and charge

- q

is projected from the origin with a horizontal speed

v

into an electric field of intensity

E

directed downward. Choose the correct statement. Neglect gravity :

A mass m carrying a charge $q$ is suspended from a string and placed in a uniform horizontal electric field of intensity $E$ . The angle made by the string with the vertical in the equilibrium position is :

Q. A charged oil drop weighing

1.6 \times 10^{- 15} N

is found to remain suspended in a uniform electric field of intensity

2 \times 10^{3} N C^{- 1}

. Find the charge on the drop.

Q. A region consists of uniform electric field

E

and uniform gravitational field

g

perpendicular to each other. A particle of charge

Q

and mass

m

begins to move in this region Deduce an equation for its trajectory.

Join BYJU'S Learning Program

An oil drop of mass m and charge +q is balanced in vacuum by a uniform electric field of intensity E The direction of this field should be

The correct option is A vertically up.mg=qEE should be upward to balance downward mg.

An oil drop of mass m and charge $+ q$ is balanced in vacuum by a uniform electric field of intensity $E$ The direction of this field should be

The correct option is A vertically up.
$m g = q E$
$E$ should be upward to balance downward $m g$ .