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

Step-by-step solution

QV ℓ = 4 3 π r 3 ρ g When the oil drop is falling freely under the effect of gravity is a viscous medium with terminal speed v , then m g = 6 π η r v ...(i) To move the oil drop upward with terminal velocity v if E is the electric field intensity applied, the E q = m g + 6 π η r v = m g + m g = 2 m g