(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider a uniformy charged thin-walled right circular cylindrical shell having total charge Q, radius R, and height h. Determine the electric field at a point a distance d from the right side of the cylinder (treat the cylinder as a collection of ring charges). Consider now a solid cylinder with the same dimenstions carrying the same charge,uniformly distrubted through its volume (treat the cylinder as a collection of disk charges)

2. Relevant equations

Ring Electric Field: [tex]\int \frac{kx dq}{(x^{2} + a^{2})^{\frac{3}{2}}} [/tex]

Disk Electric Field: [tex]\int \frac{kx\pi\sigma 2rdr}{(x^{2} + a^{2})^{\frac{3}{2}}}[/tex]

The disk equation is derived from the ring equation by treating a disk as a series of rings of infinitesimal radius and integrating using the substition [tex]dq = \pi\sigma 2rdr[/tex]

3. The attempt at a solution

For the first portion of the problem I started with the Ring Equation and used the equation [tex]dq = \lambda dx[/tex]. This gave me

[tex]k\lambda\int \frac{xdx}{(x^{2} + R^{2})^{\frac{3}{2}}}[/tex]

Using the substitution [tex] u = (x^{2} + R^{2}), du = 2xdx [/tex] I have

[tex]\frac{k\lambda}{2}\int \frac{du}{u^{\frac{3}{2}}} = \frac{-k\lambda}{\sqrt{u}}|^{d+h}_{d} = k\lambda(\frac{1}{\sqrt{d^{2} + R^{2}}} - \frac{1}{\sqrt{(d+h)^{2} + R^{2}}})[/tex]

I went online to compare my results with others and found someone had asked the https://www.physicsforums.com/showthread.php?t=188011" as myself. My answer seems to follow their line of logic. However I can't get an appropriate answer for the 2nd part of the problem. It seems to me like I'd just take the same approach and just plug [tex] dq = \rho dV =\rho\pi r^{2}dx [/tex] into the Disk equation. However they replaced [tex] dq [/tex] already, and if I go back and plug my [tex] dq [/tex] into the ring equation (like the book did with it's [tex]dq[/tex]) then I just end up with the same equation as above. Specifically I get

[tex]\rho k\pi R^{2}(\frac{1}{\sqrt{d^{2} + R^{2}}} - \frac{1}{\sqrt{(d+h)^{2} + R^{2}}})[/tex].

But since [tex] \rho = \frac{Q}{V} = \frac{Q}{\pi R^{2} h} [/tex] and [tex] \lambda = \frac{Q}{h} [/tex] both equations reduce down to

[tex]\frac{kQ}{h}(\frac{1}{\sqrt{d^{2} + R^{2}}} - \frac{1}{\sqrt{(d+h)^{2} + R^{2}}})[/tex].

Any help is greatly appreciated

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# Homework Help: Electric Field due to a charged hallow cylinder/solid cylinder on a point

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