A charged dust particle of radius `5times10^(-7)` m is moving in a horizontal electric field of intensity `6.28times10^(5)V/m` .The surrounding medium is air with coefficient of viscosity `eta=1.6times10^(-5)N-s/m^(2).` If this particle is moving with a uniform horizontal speed of `0.02m/s` ,the number of excess electrons on the drop are `6k` .Find the value of `k`.

A charged dust particle of radius 5 xx 10^(-7)m is located in a horizontal electric field having an intensity of 6.28 xx 10^(5) V m^(-1) . The surrounding medium in air with coefficient of viscosity eta = 1.6 xx 10^(-15) Ns m^(-2) . If this particle moves with a uniform horizontal speed of 0.01 ms^(-1), the number of electrons on it will be

A charged particle of radius 5xx10^(-7)m is located in a horizontal electric field of intensity 6.28xx10^(5)Vm^(-1) . The surrounding medium has the coeffecient of viscosity eta=1.6xx10^(-5)Nsm^(-2) . The particle starts moving under the effect of electric field and finally attains a uniform horizontal speed of 0.02 ms^(-1) . Find the number of electrons on it. Assume gravity free space.

A charged oil drop weighing 1.6 xx 10^(-15) N is found to remain suspended in a uniform electric field of intensity 2 xx 10^3NC^(-1) . Find the charge on the drop.

A charged particle having drift velocity of 7.5 x 10^-4 m/s in an electric field of 3 x 10^-10 V/m has a mobility in m^2 V^-1 s^-1

A charged particle of charge 1 mC and mass 2g is moving with a speed of 5m/s in a uniform magnetic field of 0.5 tesla. Find the maximum acceleration of the charged particle

A charged particle of charge 1 mC and mass 2g is moving with a speed of 5m/s in a uniform magnetic field of 0.5 tesla. Find the maximum acceleration of the charged particle

A particle is moving in a circle of radius 1 m with speed varying with time as v=(2t)m//s . In first 2 s

A particle of mass 2 xx 10^(-3) kg, charge 4 xx 10^(-3)C enters in an electric field of 5 V//m , then its kinetic energy after 10 s is

An oil drop of diameter 4 xx 10^(-6) m falls through air. iF the densities of the oil and air are 900 kg/m^(3) and 1.293 kg/m^(3) respectively and the coefficient of viscosity of air is 2.0 xx 10^(-5) Ns//m^(2) , then the terminal velocity of oil drop will be

A proton is moving with a velocity of 5 xx 10^(5) m//s along the Y-direction. It is acted upon by an electric field of intensity 10^(5) V/m along the X-direction and a magnetic field of 1 Wb//m^(2) along the Z-direction. Then the Lorentz force on the particle is :