Calculating Electric Field of Spherical Charge Distribution

[tome101]

Homework Statement

Compute the electric field generated by a spherically symmetric charged sphere of radius R with charge density of $\rho = kr^{2}$

Homework Equations

$\oint _S \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$

The Attempt at a Solution

I know that this question involves the application of Gauss' law but I don't really know how? To be honest I'm a bit sketchy on applying Gauss' law to any question. Any help would really be appreciated.

 

[ideasrule]

Imagine a Gaussian sphere of radius r, centered on the actual sphere of charge. What would the surface integral of E*dA be, in terms of r? Remember that due to symmetry, the electric field has to be constant for constant r, and must be entirely radial.

Using integration, can you also find Q_enclosed for this Gaussian sphere?

 

[tome101]

OK, by integration I've found the charge enclosed by the sphere to be (4pi*k*r^5)/5, but I'm not really sure where to go from here?

From
$\oint _S \vec{E} \cdot \vec{dA} = \frac{Q_{enclosed}}{\epsilon_0}$
I can see that I need to divide the charge enclosed by epsilon 0 then equate to the surface integral of E*da, but I'm not really sure how to calculate the surface integral of E*da?
Thanks

 

[tome101]

Ok, I've now been told that the surface integral of E*dA in this case goes to E(4pi*r^2) but I'm still not totally sure why.

 

[rwisz]

Dear student,

Thank you for your question. The solution to this problem involves using Gauss' law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. In this case, we can use a spherical Gaussian surface with radius r, centered at the origin, to calculate the electric field at any point outside the charged sphere.

To apply Gauss' law, we first need to determine the charge enclosed by our Gaussian surface. In this case, we can use the given charge density, \rho = kr^{2}, to find the total charge within the sphere. We can do this by integrating the charge density over the volume of the sphere:

Q_{enclosed} = \int_{0}^{R} \rho 4\pi r^2 dr = \frac{4}{5}\pi kR^5

Next, we can calculate the electric flux through the Gaussian surface by taking the dot product of the electric field, \vec{E}, and the differential area element, \vec{dA}. Since the electric field is spherically symmetric, we know that it must be perpendicular to the surface at every point, and thus the dot product simplifies to \vec{E}\cdot \vec{dA} = E dA \cos{\theta}, where \theta is the angle between the electric field and the surface normal.

Using the symmetry of our system, we can see that the electric field must be constant and pointing radially outward at every point on the surface. Therefore, the angle \theta is always 0 and the cosine term is equal to 1. This simplifies our equation to E dA, which is simply the magnitude of the electric field multiplied by the surface area of the Gaussian sphere, 4\pi r^2.

Now, we can set up our equation using Gauss' law:

\oint _S \vec{E} \cdot \vec{dA} = E \int_{0}^{R} 4\pi r^2 dr = \frac{4}{5}\pi kR^5 / \epsilon_0

Solving for the electric field, we get:

E = \frac{kR^3}{5\epsilon_0}

Therefore, the electric field at any point outside the charged sphere is given by this equation. This result makes intuitive sense, as we would expect the electric field