Charged Conducting Sphere

A conductor in the shape of a sphere. Due to being a conductor, assumed to be in Electrostatic Equilibrium, i.e. ${\vec{E}}_{inside}=0$

Electric Field §

Assum #TODO Trivial with [Gauss' Law](Gauss'%20Law.md)

$$\vec{E}=\{\begin{array}{ll}\frac{k_{e}q}{r^2},& r\ge r_{sphere}\\ 0,& r<r_{sphere}\end{array}$$

Electric Potential §

$$V=\{\begin{array}{ll}\frac{k_{e}q}{r},& r\ge r_{sphere}\\ 0,& r<r_{sphere}\end{array}$$

Derivations §

Electrical Potentials §

Assuming we know the Electric Field, $\vec{E}$, electrical potential is given as the line integral of the electric field:

$$\Delta V=-{\int }_i^j\vec{E}\cdot d\vec{L}$$

We define $V_f(\infty )=0$ to be our base potential. Furthermore, to go from $r_i$ to $\infty$, we take a straight line path, so $d\vec{L}$ = $d\vec{r}$

$$\Delta V=-{\int }_{r_i}^{\infty }\vec{E}\cdot d\vec{r}$$

Assuming $r_i>r_{sphere}$

=-\int {r{i}}^\infty \vec{E} , dr = -k_{e}q\int_{r_{i}}^{\infty} {\dfrac{1}{r^2}} , dr

= k_{e}q[\dfrac{1}{\infty} - \dfrac{1}{r_{i}}]

V_{f}-V_{i}=k_{e}q[\dfrac{1}{\infty}-\dfrac{1}{r_{i}}]

-V_{i} = -\dfrac{k_{e}q}{r}

$$---$$

V = \dfrac{k_{e}q}{r}

$$---$$