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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jun 2008 22:41:24 IST
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Find the electric field and potential at the center of a hemisphere of radiusR
charged uniformly with a charge of surface density sigma?
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13 Jun 2008 22:49:57 IST
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using gauss law
E*s = Q / eo
Q = sigma * 2/3 pie*R^3
E*2pie R^2 = 2/3pie*R^3 / eo
E = sigma * R / 3eo
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13 Jun 2008 22:51:19 IST
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E=dv/dr
dv = Edr
integrate
V = S sigma *R / 3eo dr
= sigma / 3eo * R^2/2
= sigma R^2 / 6eo
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14 Jun 2008 07:20:00 IST
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![\mbox{Radius of elementary ring }= R \sin \theta \\ \\<br/>Therefore \ , \ dq= \sigma * 2 \pi R \sin \theta dr \\ \\<br/>where \ dr = d\theta * R [\mbox{Use angle = arc/radius}] \\ \\<br/>dq = \sigma * 2 \pi R \sin \theta d\theta * R\\ \\<br/>\mbox{From symmetry of problem }E_x=0 \\ \\<br/>Now , dE_y = \frac{K dq}{R^2} \cos \theta \\ \\<br/>= \frac{\sigma}{2 \epsilon _0} \sin \theta \cos \theta d\theta \\ \\<br/>\mbox {On integrating the above expression from }0 \ to \ \frac{\pi}{2} \\ \\<br/>Answer= \frac{\sigma}{4 \epsilon _0 }\\ \\](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/d/5/8d5cbb54ab3282821ed6bbbdaacff16b7770a1f3.gif)
![POTENTIAL-> \\ \\<br/>dV= \frac{K dq}{R} \ ........ [Use \ V_{ring}=\frac{Kq}{\sqrt{r^2+x^2}} ]\\ \\<br/>Hence, \ dV=\frac{\sigma R}{2 \epsilon _0}\sin \theta d\theta \\ \\<br/>\mbox{Integrate as previously } \\ \\<br/>V=\frac{\sigma R}{2 \epsilon _0}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/3/d/5/3d543d198e398c01d479ba707b12a934b27f4d00.gif)
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Impossible To be Impossible is Impossible |
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14 Jun 2008 11:10:10 IST
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But answer is sigma / 4apsilon not
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16 Jun 2008 10:49:34 IST
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Can anyone prove this using Gauss's Law
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16 Jun 2008 10:58:23 IST
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hey anyone please do it yaar
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16 Jun 2008 11:44:46 IST
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see ganesha's post (1st one) .he has solved it using gauss law.
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