An oil drop, carrying six electronic charges and having a mass of `1.6xx10^(-12)g`, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving down ward with ? Ignore buoyancy.

An oil drop carrying a charge q has a mass m kg. it is falling freely in air with terminal speed v. the electric field required to make, the drop move upwards with the same speed is

An oil drop carrying a charge q has a mass m kg. it is falling freely in air with terminal speed v. the electric field required to make, the drop move upwards with the same speed is

An oil drop carrying a charge q has a mass m kg. If is falling freely in air with terminal speed v. The electric field required to make the drop move upwards with the same speed 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

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 has a charge -9.6xx10^(-19) C and mass 1.6xx10^(-15) gm. When allowed to fall, due to air resistance force it attains a constant velocity. Then if a uniform electric field is to be applied vertically to make the oil drop ascend up with the same constant speed, which of the following are correct. (g=10 ms^(-2)) (Assume that the magnitude of resistance force is same in both the cases)

An oil drop has a charge -9.6xx10^(-19) C and mass 1.6xx10^(-15) gm. When allowed to fall, due to air resistance force it attains a constant velocity. Then if a uniform electric field is to be applied vertically to make the oil drop ascend up with the same constant speed, which of the following are correct. (g=10 ms^(-2)) (Assume that the magnitude of resistance force is same in both the cases)

A charged oil drop falls with terminal velocity v_(0) in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3q, it starts moving upwards with velocity v_(0) . The initial charge on the drop is

A charged oil drop falls with terminal velocity v_(0) in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3q, it starts moving upwards with velocity v_(0) . The initial charge on the drop is

A charged oil drop fall with a terminal velocity V in the absence of elcectric field . An electric field E keep keep the oil drop stationary in it . When the drop acquire a charge 'q' it moves up with same velocity. Find the initial charge on the drop.