A water drop of mass 3.2×10−18 kg and carrying a charge of 1.6×10−19C is suspended stationary between two plates of an electric field. Given g=10m/s2, the intensity of the electric field required is
2V/m
200V/m
20V/m
2000V/m
A
2V/m
B
200V/m
C
20V/m
D
2000V/m
Open in App
Solution
Verified by Toppr
Force in electric field F=qE and F=mg (By gravitation) So,mg=qE (for equilibrium)
E=mgq
=3.2×10−18×101.6×10−19
=200V/m
Was this answer helpful?
2
Similar Questions
Q1
A water drop of mass 3.2×10−18 kg and carrying a charge of 1.6×10−19C is suspended stationary between two plates of an electric field. Given g=10m/s2, the intensity of the electric field required is
View Solution
Q2
Two charges +3.2×10−19 and −3.2×10−19 C placed 2.4˙A apart form an electric dipole. It is placed in a uniform electric field of intensity 4×105 volt/m. The electric dipole moment is
View Solution
Q3
The intensity of the electric field required to keep a water drop of radius 10−5 cm, just suspended in air when charged with one electron is approximately (g = 10newton / kg, e=1.6×10−19 coulomb)
View Solution
Q4
A water drop of radius 10−6 m is charged with one electron. The electric field required to keep it stationary is (given density of water ρ=1000kg/m3;g=9.8m/s2)
View Solution
Q5
In a Millikans oil drop experiment, an oil drop of mass 0.64×10−14kg, carrying a charge 1.6×10−19C remains stationary between two plates separated by a distance of 5 mm. Given g=9.8m/s2; the voltage that must be applied between the plates being