Calculating Charge Density and Total Charge Using Gauss's Law

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SUMMARY

The discussion focuses on calculating charge density and total charge using Gauss's Law, given an electric field \(\vec{E} = kr^{3} \hat{r}\). The charge density \(\rho\) is derived as \(\rho = 5k\epsilon_{0}r^{2}\) using the divergence of the electric field in spherical coordinates. The total charge \(Q\) within a sphere of radius \(R\) is calculated through volume integration, yielding \(Q = 4\pi \epsilon_{0} R^{5}\), and verified using the integral form of Gauss's Law, resulting in \(Q = 4\pi \epsilon_{0} kR^{5}\).

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Homework Statement



Suppose the electric field in some region is found to be \vec{E} = kr^{3} \hat{r}, in spherical coordinates (k is some constant).

(a) Find the charge density ρ.

(b) Find the total charge contained in a sphere of radius R, centered at the origin. (Do it two different ways.)

Homework Equations



The Attempt at a Solution



(a) Use the formula for the divergence of a vector in spherical basis to get \nabla . \vec{E} = 5kr^{2} so that \rho = 5k\epsilon_{0}r^{2}.

(b) Q is the volume integral of \rho over the volume of the sphere. So, we integrate over d\phi, integrate sinθ over dθ, integrate the \rho times r^{2} over dr and multiply the three results. The process gives us 4\pi \epsilon_{0} R^{5}.

Q can also be found using the integral form of Gauss's law, where the surface integral of the electric field is taken with the infinitesimal area, which is R2 sinθ dθ d∅ r, where r is the unit vector in the radial direction. We take the constants out of the integral and integrate 1 over phi and sinθ over theta to obtain the same Q as above.

Please could you check if the process and the answers are correct?
 
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a) Looks good to me. I get the same.
b) Is it a hollow spherical shell. Then you can use that the electric field is constant over the surface(fixed r) and only multiply by the area of the sphere(shell) because:
\oint \vec{E}\cdot\vec{da} = \oint E\,\text{d}a = E\oint 1\text{d}a = E 4\pi R^2 = Q/\epsilon_0 \Leftrightarrow Q = 4\pi \epsilon_0 kR^5
The dot product \vec{E}\cdot\vec{da} is just da times E cause they point in the same direction.
 

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