Electrical field outside spherical shell

[glederfein]

Homework Statement

An insulated spherical shell with its center in the origin and radius R has a surface charge density of $$\sigma(\theta)=\sigma_0\sin(\theta)$$ when $$\theta$$ is the angle from the z axis. Calculate the electrical field outside the shell at point z=R.

Homework Equations

Gauss's Law
$$\int\int{\vec{E}\cdot\vec{dS}}=4\pi{Q_{in}}$$

The Attempt at a Solution

I tried first calculating the overall charge of the spherical shell:
$$Q=R^2\int_0^{2\pi}\sigma_0\sin^2\theta{d\theta}\int_0^\pi{d\phi}=\frac{R^2\sigma_0}{2}\int_0^{2\pi}(1-\cos{2\theta})d\theta\cdot\pi=\frac{1}{2}\pi{R^2}\sigma_0\cdot{2\pi}=\pi^2R^2\sigma_0$$
However, I don't see how Gauss's Law can help me find the electrical field at that specific point.

Please help!

 

[glederfein]

anyone?

Can someone please help??

Perhaps I should somehow calculate $$\int{\frac{dq}{r^2}}$$ ?

 

[Moetasim]

Your attempt at calculating the overall charge of the spherical shell is correct. However, Gauss's Law can indeed help you find the electrical field at a specific point outside the shell.

Gauss's Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of the medium. In this case, we can consider a Gaussian surface in the shape of a sphere with radius r>R, centered at the origin.

Using the symmetry of the problem, we can see that the electric field is uniform and radial, pointing away from the center of the sphere. Hence, the electric field at point z=R will be equal to the electric field at the surface of the sphere, which is given by the electric flux through the Gaussian surface divided by the surface area of the sphere.

We can calculate the electric flux through the Gaussian surface by using the fact that the electric field is perpendicular to the surface at every point. Thus, we have:

\int\int{\vec{E}\cdot\vec{dS}}=E\cdot4\pi{r^2}

And since the enclosed charge is equal to the overall charge of the spherical shell, we have:

\int\int{\vec{E}\cdot\vec{dS}}=\frac{\pi^2R^2\sigma_0}{\epsilon_0}

Equating the two equations and solving for E, we get:

E=\frac{\pi\sigma_0R}{2\epsilon_0}

Therefore, the electric field at point z=R is given by E=\frac{\pi\sigma_0R}{2\epsilon_0}, where \epsilon_0 is the permittivity of the medium.