Potential at centre of cylinder

In summary, the problem asks for a solution to the integral ∫V(r,θ)dθ, which is a problem for which Gauss's law does not suffice.
  • #36
Pranav-Arora said:
I don't have any reasons but if I do that, I get the right answer but still, the whole thing went over my head. I never knew it would involve this heavy maths.

Thank you! :)

Well, you don't have to thank me, it was haruspex who gave you the formula.

However, I am not entirely sure you should be allowed to use this formula. TSny has tried to help you get this formula through a physical argument, you should try and complete the derivation. Otherwise you should know that the property is called a mean value theorem for harmonic functions and can be proven purely mathematically.
 
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  • #37
voko said:
Well, you don't have to thank me, it was haruspex who gave you the formula.

However, I am not entirely sure you should be allowed to use this formula. TSny has tried to help you get this formula through a physical argument, you should try and complete the derivation. Otherwise you should know that the property is called a mean value theorem for harmonic functions and can be proven purely mathematically.

I myself don't know if I am allowed to use that formula.

I tried TSny suggestion but I don't know how to solve
[tex]\int -\frac{∂V}{∂r}rhd\theta=0[/tex]
:confused:
 
  • #38
Can you take r and h out of that integral? Can you then switch the order of integration and differentiation?
 
  • #39
voko said:
Can you take r and h out of that integral?
Yes.
Can you then switch the order of integration and differentiation?
No. :(

I am ready to understand the required mathematics (if it isn't too much like those heavy ones, I do know basic calculus) if you can give me a link. :)
 
  • #40
Pranav-Arora said:
No. :(

You can justify interchanging the order of integrating and differentiation. Just think of replacing the partial derivative with respect to r by a finite difference ##\int_0^{2\pi} \frac{V(r+Δr,\theta)-V(r,\theta)}{Δr}\,d\theta##. Split into two integrals over θ, and then let Δr → 0.

Or, if you don't like pulling out the derivative, you can do the following. Integrate the equation ##\int_0^{2\pi} \frac{\partial V(r,\theta)}{\partial r}\,d\theta = 0## from ##r = 0## to ##r = R## to get ## \int_0^{R} \int_0^{2\pi} \frac{\partial V(r,\theta)}{\partial r}\,d\theta dr = 0## and interchange the order of integration so that you integrate over r first.
 
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  • #41
This is basic calculus. You have an integral, where the integrand depends on a parameter (r), which is different from the integration variable. Symbolically, ## F(r) = \int f(r, \theta) d\theta ##. Then there is a theorem stating that ## \frac {dF} {dr} = \frac {d} {dr} \int f(r, \theta) d\theta = \int \frac {\partial f} {\partial r} d\theta ##, i.e., that one can switch the order of integration and differentiation.

I am afraid you won't be able to progress much in your studies of EM unless you brush up on your calculus.
 

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