Van de Graaff Problem: Solve for Electric Field Strength

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Homework Help Overview

The discussion revolves around calculating the electric field strength generated by a van de Graaff generator with a negatively charged metal sphere. The problem specifies the radius of the sphere and the charge, and participants are tasked with determining the electric field strength at a specific point outside the sphere.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the concept of a test charge to find the force and subsequently the electric field. There is uncertainty about how to correctly calculate the distance from the center of the sphere to the point of interest, with questions about whether to add the radius of the sphere to the distance from its surface.

Discussion Status

The discussion is active, with participants providing guidance on the approach to take. Some clarification has been offered regarding the calculation of distance, but there is still a request for step-by-step assistance, indicating that participants are exploring different interpretations of the problem.

Contextual Notes

Participants express confusion about the setup of the problem, particularly regarding the addition of distances and the application of formulas. There is an emphasis on needing a clearer understanding of the electric field concept in this context.

Ryo124
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A van de Graaff generator puts negative charge on a metal sphere.

Suppose the radius of the sphere is a = 6.9 cm, and the charge on the sphere is Q = -1.0×10-8 C. Determine the electric field strength at a point 1.0 cm from the surface of the sphere (outside the sphere). I've done this problem over and over and am not getting the correct answer.
I've used E = F/q and made F = kq/r^2
...from there, I don't know what to do. Please help. Thanks
 
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I think they are aking you to use a test charge (like an electron) and find the force on it at r (F = kq'q/r^2). Then find the field E at this point.
 
step-by-step please

OK, but I am still unsure what to do.. Do you add .069m and .01m for r?? Can someone "walk" me through this step-by-step PLEASE.
 
The field outside the sphere is the same as thought the entire charge were concentrated at a point at its center. So r would be 0.069 + 0.01.
 
Got it. Thanks.
 

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