Gauss' Law of Electrostatics

a) It is easy to show that the application of Gauss' law in the outer area results in a formula that is equal to Coulomb's law. The flux inside the sphere is given as $$\Psi = \int \vec{E}\cdot\mathrm{d}\vec{A} = 0$$ because no charge is included in the given area. Therefore, the electric field vanishes and the potential is constant. b) The left side of Gauss' law gives $$\Psi = \int \vec{E}\,\mathrm{d}\vec{A} = E4\pi r^2$$ The right side can be written according to $$\Psi = \frac{Q}{\varepsilon_0} = \int \frac{\varrho}{\varepsilon_0}\,\mathrm{d}V$$ By replacing the volume element $\mathrm{d}V$ with $r^2 \mathrm{d}r\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi$ This can be written as $$\Psi = \int \frac{\varrho}{\varepsilon_0}\,\mathrm{d}V = r^2 \mathrm{d}\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi$$ Solving the integral leads to $$\Psi = \frac{1}{\varepsilon_0}\frac{4}{3}\pi r^3$$ In the next step, we can equal both equations and obtain the electric flux $$E = \frac{r\varrho}{3\varepsilon_0}$$ Finally, we can substitute $\varrho$ as follows $$\varrho = \frac{Q}{\frac{4}{3}\pi r^3}$$ This results then in $$\vec{E} = \frac{Qr}{4\pi\varepsilon_0R^3}\hat{\vec{r}}$$ This means that the electric field inside a solid sphere increases linearly with the radius. The integration of the potential gives $$\varphi(r) = \frac{Q}{4\pi\varepsilon_0 R}\left(\frac{3}{2} - \frac{r^2}{2R^2}\right)$$ c) Due to symmetry reasons, the surface of a cylinder has to be taken into account. For $r > R$ it follows: $$\Psi = \int \vec{E}\cdot\mathrm{d}\vec{A} = E 2\pi r L$$ This has to be equal to $$\Psi = \frac{Q}{\varepsilon_0} = \frac{\lambda}{\varepsilon_0}L$$ Then it follows for the electric field $$\vec{E} = \frac{\lambda}{2\pi \varepsilon_0 r}\hat{\vec{r}}$$ For $r < R$ the flux can be written as $$\int \vec{E}\cdot\mathrm{d}\vec{A} = E2\pi r L = \frac{\varrho \pi r^2 L}{\varepsilon_0}$$ which leads to $$\vec{E} = \frac{\varrho r}{2\varepsilon_0}\hat{\vec{r}} = \frac{\lambda\vec{r}}{2\varepsilon_0 \pi R^2}$$ The electric potential can now be found integrating from R to infinity for the outer area and defining $\varphi(R) = 0$: $$\varphi(r) = -\frac{\lambda}{2\pi \varepsilon_0}\ln\left(\frac{r}{R}\right)$$ And for the inner part it follows: $$\varphi(r) = \frac{\lambda}{4\pi\varepsilon_0}\left(1-\frac{r^2}{R^2}\right)$$