The magnitude of electric field E required to balance an oil drop of mass m, carrying charge q is (g = acceleration due to gravity)

Question

The magnitude of electric field E required to balance an oil drop of mass m, carrying charge q is (g = acceleration due to gravity) (A)  (q/m)  (B)  (mg/q2)  (C)  mgq  (D)  (mg/q)

Tardigrade Admin · Accepted Answer

To balance the oil drop the electric force should be equal to the weight of the oil drop. ie, qE=mg or E=(mg/q)

Q. The magnitude of electric field $E$ required to balance an oil drop of mass $m$ , carrying charge $q$ is ( $g$ = acceleration due to gravity)

$\frac{q}{m}$

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

$m g q$

$\frac{m g}{q}$

To balance the oil drop the electric force should be equal to the weight of the oil drop. ie,
$qE = m g$ or $E = \frac{m g}{q}$

Q. The magnitude of electric field E required to balance an oil drop of mass m, carrying charge q is (g = acceleration due to gravity)

mq​

q2mg​

mgq

qmg​