Maximum electric field at the surface of a Van de Graaff generator

  • #1
archaic
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Homework Statement
We have a Van de Graaff generator consisting of a spherical dome on which charge is continuously deposited by a moving belt until the electric field at the surface of the dome becomes equal to the dielectric strength of air.
If the diameter of the dome ##R=38.0## cm, and the "breakdown" electric field surrounding it is ##3.00\times10^6## V/m
a) What is the maximum value of the potential of the dome in kV?
b) What is the maximum charge on the dome in micro coulombs?
Relevant Equations
N/A
I know that the potential of the sphere at its surface is ##V(a)=kQ/a##, and the electric field generated by it is ##E(a)=kQ/a^2##, which gives me ##V(a)=aE(a)##.
When the electric field at the surface is as in the question, we have ##V(a)=38.0\times10^{-2}\times3.00\times10^6=1140000\,V=1140\,kV=1.14\times10^3kV##.
And the charge is ##Q=aV(a)/k=\frac{38.0\times10^{-2}\times1.140000}{8.99\times10^9}\times10^6=48.2\,\mu C##.
Do you guys see something missing?
I am confident that the calculations are fine, and that I have the right units..
 
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  • #2
Looks good to me. [EDIT: Ah! They give you the diameter not the radius! Yet they denote the diameter with R.]
 
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  • #3
TSny said:
Looks good to me. [EDIT: Ah! They give you the diameter not the radius! Yet they denote the diameter with R.]
OH! That's actually my bad.. I paraphrased parts of the question.
Thank you very much!
 
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