Question

The liquid drop of mass 'm' has charge,What should be the magnitude of electric field E to balance this drop? (A) mg/q (B) E/m (C) mgq (D) mg/g​

Answer 1

Given:

Liquid drop has mass M with charge q.

To find:

Magnitude of Electrostatic field in order to balance the liquid drop in mid air.

Calculation:

The weight of the liquid drop will be equal and opposite to the force exerted by the electrostatic field in order to balance the drop.

Weight of the drop will be mg.

Electrostatic Force = Eq.

$\sf{\therefore weight = electrostatic \: force}$

$\sf{ = > m \times g = E \times q}$

$\sf{ = > E = \dfrac{mg}{q} }$

In order to prevent the fall of the liquid drop, the Field Intensity has to be directed upwards.

So, final answer is :

$\boxed{ \red{ \sf{E = \dfrac{mg}{q} \: \: ( \uparrow \uparrow)}}}$