
Gauss law of Electrostatics?  
asked  1737 views  1 answers.  
State Gauss law of electrostatistics. A spherical charge distribution has a volume charge density given by, [$$:] \rho= \left\{ \begin{array}{lr} \rho_0\left(1\frac{r^2}{a^2}\right) & 0 \leq r \leq a\\ 0 & r>a \end{array} \right. [/:$$] where [$:] \rho_0 [/:$] is a constant. Calculate the charge, [$:]Q[/:$], enclosed.  
GaussElectrostaticsPhysicsMathematics  
 
0
 [$$:]\oint \vec{E} \cdot d\vec{A}=\frac{q_{enc}}{\epsilon_0}[/:$$]
This is Gauss' Law. And you are given the charge density as:
[$$:]\rho = \rho_0 \frac{1r^2}{a^2}[/:$$]
Remember that charge density is defined as:
[$$:]\rho = \frac{Q}{V}=\frac{dq}{dV}[/:$$]
Hence, we have:
[$$:]dq=\rho \; dV[/:$$]
Well, we can find [$:]dV[/:$]. We are talking about a sphere here, so it will be:
[$$:]dV=4\pi r^2 dr[/:$$]
That will be the micro volume. You can draw a diagram to get this, but this is very simply the derivative of the volume of a sphere, which is [$:]\frac{4}{3}\pi r^3[/:$].
Plug everything back in to [$:]dq[/:$], including the density:
[$$:]dq = \rho_0 \frac{1r^2}{a^2} 4\pi r^2 dr=\frac{\rho_0 4\pi}{a^2}(r^2r^4) dr[/:$$]
From this, you can get [$:]q_{enc}[/:$]. [$:]q_{enc}[/:$] is just the integral. Let's start with [$:]0 \le r < a[/:$]. The charge will be:
[$$:]q_{enc} = \int dq = \frac{\rho_0 4\pi}{a^2} \int_0^r (r^2r^4) dr=\frac{\rho_0 4\pi}{a^2} \left(\frac{r^3}{3}\frac{r^5}{5}\right)[/:$$]
The thing about Gauss's law is that, in most cases, the electric field won't be dependent on the area, hence:
[$$:]\oint \vec{E} \cdot d\vec{A}=\vec{E} \cdot \oint d\vec{A}=\vec{E}\cdot\vec{A}[/:$$]
The area is very simple. It's just [$:]4\pi r^2[/:$] because it's a sphere. Therefore:
[$$:]E(4 \pi r^2)=\frac{\frac{\rho_0 4\pi}{a^2} \left(\frac{r^3}{3}\frac{r^5}{5}\right)}{\epsilon_0}[/:$$]
Simplifying gives you:
[$$:]\therefore E=\frac{\rho_0}{a^2\epsilon_0} \left(\frac{r}{3}\frac{r^3}{5}\right)[/:$$]
This is for [$:]0 \le r < a[/:$]. Similar thing for when it's greater than [$:]a[/:$], except you will need the sphere's full charge:
[$$:]q_{enc} = \int dq = \frac{\rho_0 4\pi}{a^2} \int_0^a (r^2r^4) dr=\frac{\rho_0 4\pi}{a^2} \left(\frac{a^3}{3}\frac{a^5}{5}\right)=\rho_0 4\pi \left(\frac{a}{3}\frac{a^3}{5}\right)[/:$$]
We only integrate to [$:]a[/:$] because afterwards there's no charge. That's the sphere's total enclosed charge. *However*, the Gaussian surface's area is still [$:]4 \pi r^2[/:$], of course.
Hence we have:
[$$:]E(4 \pi r^2)=\frac{\rho_0 4\pi \left(\frac{a}{3}\frac{a^3}{5}\right)}{\epsilon_0}[/:$$]
Simplified to:
[$$:]\therefore E=\frac{\rho_0}{\epsilon_0 r^2} \left(\frac{a}{3}\frac{a^3}{5}\right)[/:$$]
 
 
Add comment 
Post Your Answer Here :
Add comment