Cylindrical shell- electric potential problem

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SUMMARY

The discussion focuses on calculating the electric potential difference between the surface of a long insulating cylindrical shell and a point above it. The cylindrical shell has a radius of 6.40 cm and a linear charge density of 8.90 μC/m. The initial calculation yielded a potential of 3.89e4 V, while the correct answer is 7.78e4 V, indicating an error in the integration process. The user is advised to review the application of Gauss' Law and the relevant formulas for electric potential.

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giacomh
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A very long insulating cylindrical shell of radius 6.40 cm carries charge of linear density 8.90μC/m spread uniformly over its outer surface. What would a voltmeter read if it were connected between the surface of the cylinder and a point 4.00 cmabove the surface?

λ=dq/dr
V=k∫dq/r


V=λk*ln([r+x]/r)
V=(8.9e-6)*9e9*ln(.104/.064)
V=3.89e4


The actual answer is 7.78e4 (twice my answer). I can't figure out what's wrong with my integral to make the answer half of what it should be. Thanks!
 
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giacomh said:
A very long insulating cylindrical shell of radius 6.40 cm carries charge of linear density 8.90μC/m spread uniformly over its outer surface. What would a voltmeter read if it were connected between the surface of the cylinder and a point 4.00 cmabove the surface?

λ=dq/dr
V=k∫dq/rV=λk*ln([r+x]/r)

λ is the linear charge density, λ=dq/dL.
Check also the formulae for V. What is the electric field around a very long cylindrical shell according to Gauss' Law?

ehild
 

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