anonymous
  • anonymous
A spherical cloud of charge of radius R carries total charge Q. The charge is distributed so that its density is spherically symmetric, i.e. it is a function of the radial distance from the centre of the sphere. Explain why the “charge cloud” is equivalent to a point charge of Q Coulombs at the centre of the sphere.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
please help me
anonymous
  • anonymous
use gauss' law.
anonymous
  • anonymous
explain please

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Hero
  • Hero
xylem, this is your question? Whew, good luck buddy
anonymous
  • anonymous
that's is hard to me 2
anonymous
  • anonymous
lol its an annoying question i cant figure it out doing my head in
anonymous
  • anonymous
construct a gaussian surface surrounding the charge cloud. Gauss's law demonstrates that the charge enclosed in any surface is equivalent to a point charge.
anonymous
  • anonymous
\begin{array}l\color{#FF0000}{\text{I}}\color{#FF7F00}{\text{ }}\color{#FF7F00}{\text{s}}\color{#FFFF00}{\text{h}}\color{#00FF00}{\text{o}}\color{#0000FF}{\text{u}}\color{#6600FF}{\text{l}}\color{#8B00FF}{\text{d}}\color{#FF0000}{\text{ }}\color{#FF0000}{\text{r}}\color{#FF7F00}{\text{e}}\color{#FFFF00}{\text{a}}\color{#00FF00}{\text{l}}\color{#0000FF}{\text{l}}\color{#6600FF}{\text{y}}\color{#8B00FF}{\text{ }}\color{#FF0000}{\text{r}}\color{#FF7F00}{\text{e}}\color{#FFFF00}{\text{v}}\color{#00FF00}{\text{i}}\color{#0000FF}{\text{s}}\color{#6600FF}{\text{e}}\color{#8B00FF}{\text{ }}\color{#FF0000}{\text{m}}\color{#FF7F00}{\text{y}}\color{#FFFF00}{\text{ }}\color{#FFFF00}{\text{P}}\color{#00FF00}{\text{h}}\color{#0000FF}{\text{y}}\color{#6600FF}{\text{s}}\color{#8B00FF}{\text{i}}\color{#FF0000}{\text{c}}\color{#FF7F00}{\text{s}}\color{#FFFF00}{\text{.}}\end{array}
anonymous
  • anonymous
gl search it yahoo answer .com maybe it's on there

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