Home
Class 12
PHYSICS
In an oil drop experiment, the distance ...

In an oil drop experiment, the distance between the capacitor plates = 0.076 m, the p.d. between the plates = 5000 V and the radius of drop = `2.75 xx 10^(-6)cm`. The minimum change in velocity with the field `= 5.27 xx 10^(5) cms^(-1)`. If viscosity of air `-1.83 xx 10^(-5)` Ns `m^(-2)`, calculate the charge on the electron.

Text Solution

Verified by Experts

The correct Answer is:
`1.6 xx 10^(-19)` coulomb
Promotional Banner

Topper's Solved these Questions

  • ELECTRONS AND PHOTONS

    SL ARORA|Exercise TYPE C : Problem For Self Practice|11 Videos

  • ELECTRONS AND PHOTONS

    SL ARORA|Exercise TYPE D : Problem For Self Practice|20 Videos

  • ELECTRONS AND PHOTONS

    SL ARORA|Exercise TYPE A : Problem For Self Practice|21 Videos

  • ELECTROMAGNETIC WAVES

    SL ARORA|Exercise Problem for self-practice|21 Videos

  • ELECTROSTATIC FORCES, CHARGES AND FIELDS

    SL ARORA|Exercise Problems For Self Practice|1 Videos

Similar Questions

Explore conceptually related problems

A rain drop of radius 0.5 mm has a terminal velocity in air 2 ms^(-1) . The viscosity of air is 18 xx 10^(-5)" dyne "cm^(-2)s . Find the viscous force on the rain drop.

The terminal velocity of a rain drop is 30cm//s . If the viscosity of air is 1.8xx10^(-5)Nsm^(-2) .The radius of rain drop is

In an oil drop experiment, the charges on oil drops were found as 1.5 xx 10^(-15), 3 xx 10^(-15), 4.5 xx 10^(-15), 6.0 xx 10^(-15) . Calculate the magnitude of the charge on the electron.

In an oil drop experiment, the following data were recorded : Plate separation = 15 mm, Distance of fall = 10 mm Potential difference = 3.2 kV, Viscosity of air = 1.82 xx 10^(-5) Ns m^(-2) Sucessive times of rise of the drop = 42 s and 78 s. Radius of drop = 2.76 m xx 10^(-6) m Calculate the change in charge on the drop between the two sets of observation. To how many electrons does this correspond ?

In Millikan oil drop experiment, a charged drop of mass 1.8 xx 10^(-14) kg is stationary between its plates. The distance between its plates is 0.90 cm and potential difference is 2.0 kilo volts. The number of electrons on the drop is

If an oil drop of weight 3.2xx10^(-13) N is balanced in an electric field of 5xx10^(5) Vm^(-1) , find the charge on the oil drop.

An oil drop with charge 1.6xx10^(-18)C is balanced in air between two metallic plates 10 mm apart. Calculate the voltage required between the plates if mass of the drop is 2.4xx10^(-14)kg .

A drop of radius 2xx10^(-5) and density 1.2xx10^(3)kg//m^(3) falls through air. The viscosity of air is 1.8xx10^(-5) N s//m^(2). Neglecting buoyancy due to air, the terminal speed of the drop is

A rain drop of radius 0.3 mm falls through air with a terminal viscosity of 1 m s^(-1) . The viscosity of air is 18 xx 10^(-5) poise. The viscous force on the rain drop is

What is the displacement current between the square plate of side 1 cm is of a capacitor, if electric field between the plates is changing at the rate of 3 xx 10^(6) Vm^(-1)s^(-1) ?