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

Question

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 (A) (m g/q) (B) (2 m g/q) (C) (m g v/q2) (D) (2 m g v/q)

Tardigrade Admin · Accepted Answer

( 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 ldots text (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

Q. 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

$\frac{m g}{q}$

$\frac{2 m g}{q}$

$\frac{m gv}{q ^{2}}$

$\frac{2 m gv}{q}$

Q. 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

qmg​

q2mg​

q2mgv​