An oil drop of mass m carrying a charge +q is held stationary in air by an electric field E. The electric field is correctly given by
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $E=mg/q$ acting upwards
For an oil drop of mass $m$ and charge $+q$ held stationary in an electric field $E$:
Force equilibrium: Electric force = Weight
$q E = m g$
Therefore, the required electric field:
$E = \frac{m g}{q}$ Acting Upwards