Capacitance in Cylinder

  • Thread starter sebby_man
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  • #1
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Hi everyone

In most problems I've seen involving capacitance of a coaxial cable, there is an inner cylinder of radius a with +Q charge and and an outer cylinder of radius b with -Q charge. The voltage is calculated from a to b. However, in this problem the inner cylinder is -Q and the outer is +Q. In this case, would I still find the voltage from a to b, or would I find the voltage from b to a? The answer key suggests that the voltage is calculated from a to b, but I'm not sure why.
 

Answers and Replies

  • #2
Hi,

You should calculate the voltage in that way that produces a positive voltage. The cause is capacitance always is a positive number. If the inner cylinder has -Q charge, the integral for voltage should be made from b to a.

I apologize for my english. I hope this can be clear for you.
 
  • #3
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Ok, so if I'm finding the voltage from b to a, then I need to calculate the integral of E dr from a to b? The answer of the integral is then [q/(2*pi*height*epsilon_0)]*ln(a/b). This seems right to me (unless I've made some simple mistake inbetween), yet the book says it should be ln(b/a) instead.
 
  • #4
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ln (a/b)=-ln(b/a)

It´s only a matter of sign
 
  • #5
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Turns out I had made a simple mistake in my calculations. Instead of [q/(2*pi*height*epsilon_0)]*ln(a/b) I should have written [-q/(2*pi*height*epsilon_0)]*ln(a/b) because the charge of the inner cylinder is negative. Thanks for helping me realize my careless mistake.
 

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