Help Needed: Calculating Excess Electron Charge of a Water Droplet

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Homework Help Overview

The problem involves calculating the excess electron charge of a stationary water droplet in an electric field. The context is centered around electrostatics and forces acting on charged objects.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conditions under which the droplet remains stationary, questioning the forces acting on it, including gravity and electric forces. There is also exploration of what would happen if the droplet were to fall.

Discussion Status

The discussion is active, with participants raising questions about the assumptions of the problem and the forces involved. Some guidance has been offered regarding the need to consider gravity alongside the electric field.

Contextual Notes

Participants note the unusual scenario of a droplet remaining suspended, prompting questions about the forces at play and the implications for the calculations involved.

huomenta12
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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!
 
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Why would the droplet remain stationary and not fall to the ground like a drop of rain?
 


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?
 


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?
 


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.
 

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