An oil drop of radius r carrying a charge q remain stationary in the presence of electric field of intensity E. If the density of oil `rho` , then

An oil drop of mass m fall through air with a terminal velocity in the presence of upward electric field of intensity E . the drop carries a charge +q . R is the viscous drag force and f is the buoyancy force . Them for the motion of the drop.

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

An oil drop of radius r and density r is held stationary in a uniform vertically upwards electric field 'E'. If p_(0)=(ltp) is the density of air and e is quanta of charge, then the drop has–

An oil drop of mass m carrying charge - Q is to be held stationary in the gravitational field of the earth. What is the magnitude and direction of the electrostatic field required for this purpose ?

An oil drop of radius 10^(-6)m carries a charge equal to that on an electron. If the density of the oil is 2xx10^(3)kg m^(3) , find the electric field required to keep it stationary. Take e=1.6xx10^(-19)C and g=9.8 ms^(-2) .

An oil drop having a charge q and apparent weight W falls with a terminal speed v in absence of electric field . When electric field is switched on it starts moving up with terminal speed v. The strength of electric field is

An oil drop having a charge q and apparent weight W falls with a terminal speed v in absence of electric field . When electric field is switched on it starts moving up with terminal speed v. The strength of electric field is

A charge oil drop of charge q is falling under gravity with termined velocity v in the absence of electric filed . A electric field can keep the oil drop stationary . If the drop acquires an additional charge , it moves up with velocity 3v in that field . find the new charge on the drop.