Question

An oil drop carrying a charge of 2 electrons has a mass of 3.2 × 10−17 kg. It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is

Answer 1

Answer:

When the oil drop is falling freely under the effect of gravity is a various medium with terminal speed v, then mg = 6πηrv . . . (i)

To move the oil drop upward with terminal velocity v if E is the electric field intensity applied, the

Eq = mg + 6πηrv = mg + mg = 2mg

So E = 2 mg/q

Explanation:

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