Question

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

Answer 1

Dear Student,

◆ 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

Therefore, magnitude of electric field is mg/q.

Thanks for asking. Hope this is helpful...

Answer 2

The magnitude of electric field that balance this drop is $\dfrac{mg}{q}$.

Explanation:

Let m is the mass of the drop having charge of q. We need to find the magnitude of electric field E to balance this drop. The force acting on an object is force of gravity such that the gravitational force is balanced by the electric force as :

$qE=mg$

E is the electric field

We get the value of electric field as :

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

So, the magnitude of electric field that balance this drop is $\dfrac{mg}{q}$. Hence, this is the required solution.

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