Potential at centre of cylinder

[voko]

If V were constant, you could. It is not.

However, it is piece-wise constant along the circle.

 

[Saitama]

However, it is piece-wise constant along the circle.

Sorry but the following is completely a shot in the dark.

Do you mean something like dividing the integral into two separate integrals? One for 0 to 20 degrees and the other from 20 to 360 degrees?

 

[voko]

Sorry but the following is completely a shot in the dark.

Do you mean something like dividing the integral into two separate integrals? One for 0 to 20 degrees and the other from 20 to 360 degrees?

Any reason why you could not do that?

 

[Saitama]

Any reason why you could not do that?

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! :)

 

[haruspex]

Property 1 states that "The value of V at a point (x, y) is equal to the average value of V around this point...", does this mean $V_{axis}$ is the average of 110 V and 0 V, i.e 55.5 V?

No, it means that it is the average around the circle. Taking that circle at its max radius, it becomes the average around the cylinder. And that's what you calculated with the integral, but it's just (20*110+340*0)/360.

 

[voko]

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.

 

[Saitama]

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
\int -\frac{∂V}{∂r}rhd\theta=0

 

[voko]

Can you take r and h out of that integral? Can you then switch the order of integration and differentiation?

 

[Saitama]

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. :)

 

[TSny]

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.

 

[voko]

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.