Question

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

Options

• `(mg)/q`

• `(2mg)/q`

• `(mqv)/q`

• `(2mgv)/q`

Answer 1

`(2mg)/q`

Explanation:

When an oil drop falls freely under gravity in another medium with terminal speed Vl, mg = `6pinrv` ......(i)

To move the oil drop upward with terminoil velocity 'v', if E is the intensity of the applied electric field, the Eq = mg + `6pinrv` = mg + mg = 2mg. so, E = 2mg/q.