How many electrons would it take for a balloon to stick to the ceiling?

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SUMMARY

The discussion centers on calculating the number of electrons required for a balloon to adhere to a ceiling or wall through electrostatic forces. The key equations referenced include the charge of an electron (e = 1.60 x 10^-19 C) and the relationship between charge (q = Ne) and the number of electrons (N). Participants clarified that for a balloon to stick to a ceiling, the electrostatic force must counteract the gravitational force directly, while on a wall, friction plays a significant role in maintaining adhesion.

PREREQUISITES
  • Understanding of electrostatic forces and their calculations
  • Basic knowledge of gravitational force equations (F_g = mg)
  • Familiarity with frictional force concepts (F_{\text{counter}} = \mu \cdot N)
  • Knowledge of charge quantification (q = Ne)
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  • Research the principles of electrostatics and Coulomb's Law
  • Explore the concept of friction and its role in adhesion
  • Study gravitational force calculations in different contexts
  • Learn about the properties of materials that affect electrostatic interactions
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Homework Statement


How many electrons would it take for a typical balloon to stick onto a ceiling? State assumption for ceiling size or balloon size if necessary.


Homework Equations


e = 1.60 x 10 ^ -19 C
q = Ne


The Attempt at a Solution


I thought that it took only one electron for a balloon to induce a slight charge of opposite polarity in the surface of the ceiling. But I do not believe I am right. Please, any help would be appreciated. Thanks
 
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Hi there,

For the balloon to stick on the wall, the friction must compensate the weight of the balloon. Knowing that friction is dependent on the attraction between the wall and the balloon, you should be able to find this out.

Cheers
 
i still don't get it.. anyone else care to explain?
 
Hi there,

Let's try with a bit more details. The balloon has a weight, quite little but still:

F_g = mg

For the balloon to hold on the wall, you must counter weight this force:

\sum \vec{F} = 0

on the vertical axe (y):
-F_g + F_{\text{counter}} = 0

This counteracting force is the friction on the wall created by the electrostatic interaction.

F_{\text{counter}} = \mu \cdot N

I believe I gave more than enough information on your problem. The rest of the solution should be quite easy to find out.

Cheers
 
fatra2 said:
Hi there,


This counteracting force is the friction on the wall created by the electrostatic interaction.

F_{\text{counter}} = \mu \cdot N

But the balloon should stick to the ceiling, so friction doesn't come into it, and the electrostatic
force has to counter gravity directly.
 
Hi there,

The problem is a bit different when you stick the balloon to the ceiling. In this case, the electrostatic force should only counter the gravity directly.

But the original question was on a wall, where only the friction of the balloon can counteract the weight.

Cheers
 

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