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For calculating electric field outside a nonconducting sphere with a hollow spherical cavity. When I use the rule (Charge density= $dQ/dV$), I don't know exactly what is $dV$, is the volume here refers to the volume of the Gaussian surface ($V= 4/3 \pi r^3$) so that $dV$ will be = $ \pi r^2 dr$, or the $V$ is the volume containing the charges only, so it will be =$ V_0 – V_1 = 4/3 \pi r_0^3 - 4/3 \pi r_1^3 $. Thus, since $ r_0$ and $r_1$ are constants, therefore $dV$ will be = 0?

Note: $r_0$ is the radius for the whole sphere, $r_1$ is the radius for the cavity, and $r$ is for the Gaussian surface.

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1 Answer 1

The charge density is the charge per unit volume of the object that contains the charge. If the charge density is constant on that object, then the charge density is just the total charge of the object divided by the total volume of the object.

This case case, the volume of the object is $V_0 - V_1$, so if $Q$ denotes its total charge, then the charge density $\rho$ is given by $$ \rho = \frac{Q}{V_0-V_1}. $$

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Thnx alot, but What if it is not constant and given as a fuction for example $(ρ= (ρ_0 * r_1)/r)$ –  Abdulrahman Hessen Mar 15 '13 at 16:08
Well if the density is given to you, then there is no longer any need to calculate the this case, in your notation, you would just write $\rho = dQ/dV$ where $dQ$ here is a small amount of charge contained in $dV$ a small volume of the object. –  joshphysics Mar 15 '13 at 16:14
Unfortunately Still not clear for me. Well, here is the question as it is in the text book: An electric charge $Q$ is distributed through out a nonconducting sphere of radius $ r_0 $ and has a spherical cavity of radius & $r_1$ centered at the sphere's centre. Assume the charge $ Q $ is distributed in the shell (i.e. between $ r=r_1 $ and $ r=r_0 $) and that the charge density varies as ρ= $(ρ_0 * r_1 / r)$ Find the electric field as a fuction of $r$ for $r>r_0$ –  Abdulrahman Hessen Mar 15 '13 at 17:01
@AbdulrahmanHessen: Can you write down an expression for $dV$? You can infer that $Q_{\rm inc} = \int \rho dV$, so if you can write down good limits for the integral, over a single variable (hint: what should that variable be?), you're good to go. –  Jerry Schirmer Aug 12 '13 at 23:36

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