Question

the liquid drop of mass 'm' has a change.What should be the magnitude of electric field E to balance this drop​

Answer 1

Answer:

Answer

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Explanation:

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

Answer 2

Given:

A liquid drop of mass m has a charge q.

To find:

Magnitude of electrostatic field to balance the liquid drop.

Calculation:

Let the Electrostatic Field be E :

The liquid drop will be balanced (translational equilibrium) only when the upward directing Electrostatic force will be equal and opposite to the Gravitational Force.

$\sf{\therefore \: F_{E} = F_{g}}$

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

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

$\sf{ = > \: E = \dfrac{m g}{q}}$

The direction of the Electrostatic field will be upwards.

So, final answer is:

$\boxed{ \rm{ \: E = \dfrac{m g}{q} \: \: ( \uparrow \uparrow)}}$