
#1
Dec2610, 01:54 PM

P: 39

1. The problem statement, all variables and given/known data
I'm working out of Griffith's "Intro to Electrodynamics" and the problem states: "A conical surface (an empty icecream cone) carries a surface charge [tex]\sigma[/tex]. The height of the cone is h as is the radius of the top. Find the potential difference between points a (the vertex) and b (the center of the top). 2. Relevant equations and Attempt at a solution So, since this is the chapter that I'm in, I'm going to use [tex]\[V(R)=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\frac{da{}'}{R}\][/tex]. Now since a is at the vertex I chose [tex]\[\vec{a}=0\][/tex] and [tex]\[\vec{b}=h\hat{z}\][/tex]. Thus the equation would become [tex]\[V(\mathbf{b})V(\mathbf{a})=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{{da}'}{\sqrt{(h{z}')^2+{s}'^2}} \frac{{da}'}{\sqrt{{z}'^2+{s}'^2}}\right ]\][/tex] Now da' is what I was having a little trouble attaining, so I thought the best place to start would be with the surface area of the cone: [tex]\[a'=\pi s\sqrt{s^2+z^2}\][/tex] but since the radius s is equal to the height z in our case the formula becomes [tex]\[a'=\pi s\sqrt{s^2+s^2}=\sqrt{2}\pi s^2\][/tex]. Now since fractions of this area can be represented by multiplying in terms of the angle that determines the fraction of area, [tex]\[\frac{\theta }{2\pi }\][/tex]. Thus [tex]\[a'=(\sqrt{2}\pi s^2)\cdot (\frac{\theta }{2\pi })=\frac{\sqrt{2}}{2}s^2\theta \][/tex] and if I consider the angle to be small [tex]\[a'=\frac{\sqrt{2}}{2}s^2d\theta \][/tex]. Now to find the differential area I should subtract to get [tex]\[da'=\frac{\sqrt{2}}{2}(s+ds)^2d\theta \frac{\sqrt{2}}{2}s^2d\theta=\frac{\sqrt{2}}{2}d\theta(s^2+sds+ds^2s^2)=\frac{\sqrt{2}}{2}sdsd\theta\] [/tex] since ds^2 is to small to matter. The main equation then becomes: [tex]\[V(\mathbf{b})V(\mathbf{a})=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{{da}'}{\sqrt{(h{z}')^2+{s}'^2}} \frac{{da}'}{\sqrt{{z}'^2+{s}'^2}}\right ]=\frac{\sigma }{4\pi \varepsilon _{0}}\int _{S}\left [ \frac{1}{\sqrt{(h{s}')^2+{s}'^2}} \frac{1}{\sqrt{{s}'^2+{s}'^2}}\right ](\frac{\sqrt{2}}{2}{s}'{ds}'{d\theta}' )\][/tex] [tex]\[=\frac{\sqrt{2}\sigma }{8\pi \varepsilon _{0}}\int_{0}^{2\pi }\int_{0}^{h}\left ( \frac{{s}'}{\sqrt{(h{s}')^2+{s}'^2}}\frac{\sqrt{2}}{2} \right ){ds}'{d\theta}' \][/tex] [tex]\[=\frac{\sqrt{2}\sigma }{4\varepsilon _{0}} [(hln({s}'h)+\frac{{s}'^3}{3}+{s}')_{0}^{h}\frac{\sqrt{2}}{2}h]\][/tex] but the above does not converge when evaluated so I'm at a loss. This isn't for a class or anything, I'm just self studying so answer at your convenience. 



#2
Dec2710, 10:24 AM

Sci Advisor
HW Helper
P: 6,559

Try slicing the cone along the vertical axis into rings of area [itex]dA = 2\pi s dz[/itex] where s = radius of the ring at height z, which is a linear function of z. So each ring carries a charge that is proportional to z. That should be easy to integrate.
AM 


Register to reply 
Related Discussions  
Electric Potential of Uniformly Charged Cone  Advanced Physics Homework  12  
Electric Potential Cone  Advanced Physics Homework  2  
finding potential difference between two points on a cone  Advanced Physics Homework  3  
Potential difference/cone  Introductory Physics Homework  1  
Uniformly Charged Cone  Potential Difference  Introductory Physics Homework  5 