Determining the capacitance of the two conductors

  • #1

Homework Statement


For a given system, a conducting cylinder with radius ##r=a## with a linear charge density ##Q'## and a conducting surface at a distance ##z=h## from the cylinder, calculate the linear capacitance of the cylinder.Take that ##h>>a##
##C'=\frac{Q'}{U}##
image.jpg

Homework Equations


3. The Attempt at a Solution [/B]
This is my first thread on physicsforums so you tell me if i break any rules ok :D?
As shown in the image above, i took a mirror image of an object and charged it with a negative sign of the given charge. All i need to find is the potential from the cylinder to the image cylinder right?
The total electric field along the line i am integrating the potential (##\int_{r1}^{r2} Edl=U##) is equal to the sum of the two electric fields now present which is just two times the first one. That would give us the electric field of a cylinder ##E=\frac{Q'}{2\pi\epsilon_0r}## but times ##2## and my limits of my integration should be ##\int_{a}^{2h}##. When i integrate i get ##U=\frac{Q'}{\pi \epsilon_o}ln{\frac{2h}{a}}##. This is where my result disagrees with the result i have been given by the professor. He states ##U=\frac{Q'}{2\pi \epsilon_o}ln{\frac{2h}{a}}##. Where am i wrong?
 

Answers and Replies

  • #2
TSny
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The total electric field along the line i am integrating the potential (##\int_{r1}^{r2} Edl=U##) is equal to the sum of the two electric fields now present which is just two times the first one.
For an arbitrary point between the conductors, the field from one conductor does not equal the field from the other conductor. So, it is not a matter of just multiplying by 2.
my limits of my integration should be ##\int_{a}^{2h}##.
These limits don't seem right. How are you defining the variable of integration? How are r1 and r2 expressed in terms of your variable of integration? It will be helpful if you show the complete setup of the integral, including the integrand.
 

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