Van de Graaff generator energy question

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Homework Help Overview

The problem involves calculating the energy required to add an additional electron to a Van de Graaff generator charged to 50,000 volts. The context is within the realm of electrostatics and energy calculations related to electric potential.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between volts, coulombs, and joules, with one participant attempting a calculation based on the given voltage and charge of an electron. There is also a question regarding the sign of the voltage and the implications of adding an electron to a charged sphere.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the voltage sign and the nature of the charge on the sphere. Some guidance has been offered regarding the calculation, but there is no explicit consensus on the interpretation of the problem.

Contextual Notes

There is uncertainty regarding whether the sphere is initially positively or negatively charged, which affects the interpretation of the energy calculation. The original poster's intent and the problem's assumptions are also under scrutiny.

Aprilshowers
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I would appreciate help with this problem:
If a Van de Graaff generator is charged to 50,000 volts, how much
energy does it take to add an additional electron to the charge on the
sphere? (I know that the charge on an electron is -1.6 x 10^-19 C)
Thanks
 
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A volt is a "joule/coulomb"

You got volts, you got coulombs, you want joules.
 
So I multiplied 50000 x -1.6x10^-19
and I get for an answer: -8x10^-18 joules
Does that look right?
 
Aprilshowers said:
So I multiplied 50000 x -1.6x10^-19
and I get for an answer: -8x10^-18 joules
Does that look right?

Your exponent is not correct. If the sphere is negatively charged to begin with, to add another electron will require positive work. The original 50,000V potential of the sphere would actually be negative. I'm not sure what the intent of the problem is. "Additional electron" hints that it was negatively charged to begin with, but the positive voltage suggests it was positive.
 

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