Potential Related to Electric Field?

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Homework Help Overview

The discussion revolves around calculating the magnitude of charge required to create a specific potential difference of 4.9 kV between a carpet and a shoe, modeled as large sheets of charge. The problem is situated within the context of electrostatics and capacitance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between charge, capacitance, and potential difference, questioning how to approach the problem given the lack of specific area information for the shoe. Some suggest using charge per unit area (σ) instead of total charge. Others propose estimating the shoe's dimensions to treat it as a parallel plate capacitor.

Discussion Status

The discussion is active, with participants offering various insights and calculations related to capacitance and charge. There is no explicit consensus on the approach, but several lines of reasoning are being explored, including the estimation of capacitance based on assumed dimensions.

Contextual Notes

Participants note the absence of specific area and capacitance values, which are critical for solving the problem. The precision of the 4.9 kV figure is also questioned, suggesting a need for further exploration of its significance.

mitchellhewer
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A manufacturer claims that a carpet will not generate more than 4.9kV of static electricity. What magnitude of charge would have to be transferred between a carpet and a shoe for there to be a 4.9kV potential difference between the shoe and the carpet. Approximate the shoe and the carpet as large sheets of charge separated by a distance d = 1.0mm. Area is not given in the question. Could someone please explain how to do this question in detail? Thank you in advance.
 
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Q=CV. What's the capacitance of the shoe?
 
There is no area of the shoe given. Apparently, someone has told me that I can solve it in terms of σ, the charge per unit area, not the total charge.
 
And there is no capaicitance given either.
 
Why 4.9kV - sounds oddly precise.

The only thing I can suggest is that you need to at least estimate the size of a shoe, then treat it as a simple parallel plate capacitor.
Then C = 8.854×10−12* A/d

Each plate area A - say .1 * .05 gives C= 4.4 x 10-11.

Sounds about right - 40 to 50pF

So Q = 4900 * 50pF = 25 x 10-8 Coulombs.

You want electrons with that?
 

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