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

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Homework Help Overview

The discussion revolves around determining how many electrons are needed for a balloon to adhere to a ceiling, with considerations for the balloon's weight and the forces involved in the interaction with the ceiling. The subject area includes electrostatics and forces.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the weight of the balloon and the forces required for it to stick to a surface. Questions arise regarding the role of friction versus electrostatic forces in this context.

Discussion Status

Some participants have offered insights into the forces at play, including gravitational and electrostatic forces. There is an ongoing exploration of how these forces interact, particularly in the context of sticking to a ceiling versus a wall. Multiple interpretations of the problem are being discussed.

Contextual Notes

Assumptions regarding the size of the balloon and ceiling have not been explicitly stated, which may affect the calculations and reasoning presented. The distinction between sticking to a wall and a ceiling is also a point of contention in the discussion.

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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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