Cylindrical shell- electric potential problem

AI Thread Summary
The discussion centers 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 of the potential difference yields a result of 3.89e4 volts, which is incorrect as the expected answer is 7.78e4 volts. Participants suggest reviewing the integral used for the potential and the application of Gauss' Law to understand the electric field around the cylindrical shell. The conversation emphasizes the importance of correctly applying formulas for electric potential and charge density.
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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