Dalam Nama Bapa dan Putera dan Roh Kudus. Amin.
\section{Medan Listrik akibat Muatan Berbentuk Kulit Bola}
Andaikan di ruang $\mathbb{R}^3$ ada muatan berbentuk kulit bola
\[ S^2(R) := \{\vec{r} \in \mathbb{R}^3 ~|~ |\vec{r}| = R\} \]
yang memiliki rapat muatan luasan $\sigma \in \mathbb{R}$ yang seragam, di mana $R \in \{0\}\cup\mathbb{R}$.
Posisi titik pada $S^2(R)$ tentu saja adalah
\[ \vec{r}' := R(\hat{x}\sin\theta'\cos\phi' + \hat{y}\sin\theta'\sin\phi' + \hat{z}\cos\theta') \]
di mana $\hat{x} := (1, 0, 0)$, $\hat{y} := (0, 1, 0)$, $\hat{z} := (0, 0, 1)$, $\theta' \in [0, \pi]$, $\phi' \in \{0\}\cup(0, 2\pi)$.
Posisi sebarang titik di ruang $\mathbb{R}^3$ adalah
\[ \vec{r} := r(\hat{x}\sin\theta\cos\phi + \hat{y}\sin\theta\sin\phi + \hat{z}\cos\theta) \]
di mana $r \in \{0\}\cup\mathbb{R}$, $\theta \in [0, \pi]$, dan $\phi \in \{0\}\cup(0, 2\pi)$.
Selanjutnya, kita asumsikan $\theta = 0$.
Tentu saja,
\[ \vec{r} - \vec{r}' = -\hat{x}R\sin\theta'\cos\phi' - \hat{y}R\sin\theta'\sin\phi' + \hat{z}(r - R\cos\theta') \]
dan
\[ |\vec{r} - \vec{r}'|^2 = r^2 + R^2 - 2rR\cos\theta'. \]
Medan medan listrik di titik $\vec{r}$ akibat muatan $S^2(R)$ adalah
\[ \vec{E} = \frac{\sigma R^2}{4\pi\epsilon_0}\int_0^\pi\int_0^{2\pi} \frac{\vec{r} - \vec{r}'}{|\vec{r} - \vec{r}'|^3}\sin\theta'\,d\phi'\,d\theta'. \]
\[ \vec{E} = \frac{\sigma R^2}{2\epsilon_0}\hat{z}\int_0^\pi \frac{r - R\cos\theta'}{(r^2 + R^2 - 2rR\cos\theta')^{3/2}}\sin\theta'\,d\theta'. \]
Apabila $l^2 := r^2 + R^2 - 2rR\cos\theta'$, maka $2l\,dl = 2rR\sin\theta'\,d\theta'$, $r - R\cos\theta' = r - (r^2 + R^2 - l^2)/(2r)$, dan $\sin\theta'\,d\theta' = l\,dl/(rR)$, sehingga
\[ \vec{E} = \frac{\sigma R^2}{2\epsilon_0}\hat{z}\int_{|r - R|}^{r + R}\frac{1}{l^3}\left(r - \frac{r^2 + R^2 - l^2}{2r}\right)\frac{l}{rR}dl. \]
\[ \vec{E} = \frac{\sigma R}{2\epsilon_0r}\hat{z}\int_{|r - R|}^{r + R} \left[\left(r - \frac{r^2 + R^2}{2r}\right)l^{-2} + \frac{1}{2r}\right]dl. \]
\[ \vec{E} = \frac{\sigma R}{2\epsilon_0r}\hat{z}\left.\left[\left(\frac{r^2 + R^2}{2r} - r\right)l^{-1} + \frac{1}{2r}l\right]\right|_{l = |r - R|}^{l = r + R}. \]
\[ \vec{E} = \frac{\sigma R}{2\epsilon_0r}\hat{z}\left[\left(\frac{r^2 + R^2}{2r} - r\right)\left(\frac{1}{r + R} - \frac{1}{|r - R|}\right) + \frac{1}{2r}(r + R - |r - R|)\right]. \]
Apabila $r < R$, maka $|r - R| = R - r$, sehingga
\[ \vec{E} = \frac{\sigma R}{2\epsilon_0r}\hat{z}\left[\left(\frac{r^2 + R^2}{2r} - r\right)\left(\frac{1}{r + R} - \frac{1}{R - r}\right) + \frac{1}{2r}(r + R - (R - r))\right] = \vec{0}. \]
Apabila $r > R$, maka $|r - R| = r - R$, sehingga
\[ \vec{E} = \frac{\sigma R}{2\epsilon_0r}\hat{z}\left[\left(\frac{r^2 + R^2}{2r} - r\right)\left(\frac{1}{r + R} - \frac{1}{r - R}\right) + \frac{1}{2r}(r + R - (r - R))\right] \]
alias
\[ \vec{E} = \frac{\sigma}{\epsilon_0}\hat{z}\frac{R^2}{r^2}. \]
Karena $\sigma = Q/(4\pi R^2)$, di mana $Q$ adalah muatan total dari $S^2(R)$, maka
\[ \vec{E} = \frac{Q}{4\pi R^2}\frac{1}{\epsilon_0}\hat{z}\frac{R^2}{r^2} = \frac{Q}{4\pi\epsilon_0r^2}\hat{z} \]
sesuai yang diharapkan.
Om Swastyastu.
