Excess electrons needed for a water droplet to levitate in Earth's electric field

[huomenta12]

Hello everybody,

Here is a question that I had problems with in my textbook. I can't figure it out.Can you help me out?

A water droplet of radius 0.0018mm remains stationary in the air.If the electric field going downwards of the Earth is 150 N/C,how many excess electron charges must the water droplet have?


Relevant equations may be :

F= kQ₁Q₂ / r² for force


k= 9.0 x 10⁹ Nm²/C²

E = F / q for electric field

Electron charge = 1.6 x 10^-19C
Electron Mass = 9.11 x 10^-31kg

Thanks to anyone who can help!

 

[kuruman]

Why would the droplet remain stationary and not fall to the ground like a drop of rain?

 

[huomenta12]

I don't know. Thats what I thought first, but the question is like how I wrote it. What would be the method if it did fall like a normal water drop?

 

[kuruman]

You are told that the droplet is suspended in mid air and you should believe that. How many forces do you think act on the drop when it is suspended in mid air?

 

[huomenta12]

Well, there is the force of gravity and the force of the electron that goes the opposite direction to the electric field. I forgot totally about the gravity! Ok I'll try to calculate now.