Question

the liquid drop of mass m has a charge q.what should be the magnitude of electric field E to balance this drop?

Answer 1

Answer:

(A) E = mg/q

● Explaination -

Consider a liquid drop of mass m and charge q.

For electric field to balance liquid drop, gravitational force and electrostatic force should be equal.

Fg = Fe

mg = qE

E = mg/q

Answer 2

The liquid drop of mass m has a charge q.

To find : The magnitude of electric field to the balance drop.

solution : mass of liquid drop = m

charge on liquid drop = q

electric field = E

electric force acting on the drop is balanced by its weight.

⇒ electric force = weight of drop

⇒charge on drop × electric field = mass of drop × g

⇒qE = mg

⇒E = mg/q

Therefore the Electric field to balance the liquid drop is mg/q .