A charged water drop whose radius is $0.1 μ m$ is in equilibrium in the electric field. If charge on it is equal to that of an electron, then intensity of an electric field will be___N/C.

1.61

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

262

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

26.2

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

1610

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

Open in App

Solution

The correct option is B $262$
Given,

$r = 0.1 μ m = 10^{- 7} m$

$g = 10 m / s^{2}$

$e = 1.6 \times 10^{- 19} C$

$ρ = 1000 k g / m^{3}$

Volume of the water drop,

$V = \frac{4}{3} π r^{3} = \frac{4}{3} \times 3.14 \times 10^{- 21} = 4.186 \times 10^{- 21} m^{3}$

$m = ρ . V = 1000 \times 4.186 \times 10^{- 21} = 4.186 \times 10^{- 18} k g$

At equilibrium, $e E = m g$

$E = \frac{m g}{e}$

$E = \frac{4.186 \times 10^{- 18} \times 10}{1.6 \times 10^{- 19}} = 262 N / C$

The correct option is B.

Suggest Corrections

Similar questions

Q. A charged water drop of radii

0.1 μ m

under equilibrium in some electric field. The charge on the drop is equivalent to electronic charge. The intensity of electric field is:

Q. A charged water drop of radius r is in the equilibrium in an electric field. If charge on it equal to charge on an electron, then intensity field will be: (density of water = d, assume gravity)

Q. There are n electrons of charge e on a drop of oil of density

ρ

. It is in equilibrium in an electric field E. Then radius of drop is

There are $^{'} n^{'}$ electrons of charge $^{'} e^{'}$ in a drop of oil of density $ρ$ . It is in equilibrium in an electric field $E$ . Then the radius of drop is :

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

Join BYJU'S Learning Program

A charged water drop whose radius is 0.1μm is in equilibrium in the electric field. If charge on it is equal to that of an electron, then intensity of an electric field will be___N/C.

The correct option is B 262Given,r=0.1μm=10−7mg=10m/s2e=1.6×10−19Cρ=1000kg/m3Volume of the water drop,V=43πr3=43×3.14×10−21=4.186×10−21m3m=ρ.V=1000×4.186×10−21=4.186×10−18kgAt equilibrium, eE=mgE=mgeE=4.186×10−18×101.6×10−19=262N/CThe correct option is B.

A charged water drop whose radius is $0.1 μ m$ is in equilibrium in the electric field. If charge on it is equal to that of an electron, then intensity of an electric field will be___N/C.