Andaikan di ruang $\mathbb{R}^3$ ada muatan dengan distribusi berbentuk penggal garis terhingga dengan rapat muatan $\lambda \in \mathbb{R}$, yaitu
\[ L(a, b) := \{(0, 0, z') ~|~ a < z' < b\} \]
di mana $a, b \in \mathbb{R}$ dan $a < b$.
Posisi titik pada $L(a, b)$ adalah $\vec{r}' := z'\hat{z}$ di mana $\hat{z} := (0, 0, 1)$.
Posisi sebarang titik pada ruang $\mathbb{R}^3$ adalah
\[ \vec{r} := \hat{x}l\cos\phi + \hat{y}l\sin\phi + \hat{z}z \]
di mana $\hat{x} := (1, 0, 0)$, $\hat{y} := (0, 1, 0)$, $l \in \mathbb{R}^+$, $\phi \in \{0\}\cup(0, 2\pi)$, dan $z \in \mathbb{R}$.
Medan listrik yang terjadi di titik $\vec{r}$ tentu saja adalah
\[ \vec{E} = \frac{\lambda}{4\pi\epsilon_0}\int_a^b \frac{\vec{r} - \vec{r}'}{|\vec{r} - \vec{r}'|^3}dz' \]
di mana $\epsilon_0$ merupakan permitivitas listrik dalam ruang hampa.
Tentu saja,
\[ \vec{E} = \frac{\lambda}{4\pi\epsilon_0}\int_a^b \frac{\hat{x}l\cos\phi + \hat{y}l\sin\phi + \hat{z}(z - z')}{[l^2 + (z - z')^2]^{3/2}}dz' = E_x\hat{x} + E_y\hat{y} + E_z\hat{z}. \]
Tentu saja,
\[ E_x = \frac{\lambda}{4\pi\epsilon_0}(l\cos\phi)\int_a^b \frac{dz'}{[l^2 + (z - z')^2]^{3/2}}. \]
Dengan substitusi $z - z' = l\tan\alpha$, maka diperoleh $-dz' = l\sec^2\alpha\,d\alpha$. Nilai $\alpha_a$ dan $\alpha_b$ didefinisikan sedemikian $z - a = l\tan\alpha_a$ dan $z - b = l\tan\alpha_b$, sehingga
\[ E_x = \frac{\lambda}{4\pi\epsilon_0}(l\cos\phi)\int_{\alpha_a}^{\alpha_b} \frac{-l\sec^2\alpha\,d\alpha}{l^3\sec^3\alpha}. \]
\[ E_x = -\frac{\lambda}{4\pi\epsilon_0l}(\cos\phi)\int_{\alpha_a}^{\alpha_b} \cos\alpha\,d\alpha. \]
\[ E_x = \frac{\lambda}{4\pi\epsilon_0l}(\cos\phi)(\sin\alpha_a - \sin\alpha_b). \]
\[ E_x = \frac{\lambda}{4\pi\epsilon_0l}\left(\frac{z - a}{\sqrt{l^2 + (z - a)^2}} - \frac{z - b}{\sqrt{l^2 + (z - b)^2}}\right)\cos\phi. \]
Dengan penalaran yang sama, diperoleh
\[ E_y = \frac{\lambda}{4\pi\epsilon_0l}\left(\frac{z - a}{\sqrt{l^2 + (z - a)^2}} - \frac{z - b}{\sqrt{l^2 + (z - b)^2}}\right)\sin\phi. \]
Komponen yang lain adalah
\[ E_z = \frac{\lambda}{4\pi\epsilon_0}\int_a^b \frac{z - z'}{[l^2 + (z - z')^2]^{3/2}}dz'. \]
\[ E_z = \frac{\lambda}{4\pi\epsilon_0}\int_{\alpha_a}^{\alpha_b} \frac{l\tan\alpha}{l^3\sec^3\alpha}(-1)l\sec^2\alpha\,d\alpha. \]
\[ E_z = -\frac{\lambda}{4\pi\epsilon_0l}\int_{\alpha_a}^{\alpha_b} \sin\alpha\,d\alpha. \]
\[ E_z = \frac{\lambda}{4\pi\epsilon_0l}(\cos\alpha_b - \cos\alpha_a). \]
\[ E_z = \frac{\lambda}{4\pi\epsilon_0l}\left(\frac{l}{\sqrt{l^2 + (z - b)^2}} - \frac{l}{\sqrt{l^2 + (z - a)^2}}\right). \]
\[ E_z = \frac{\lambda}{4\pi\epsilon_0}\left(\frac{1}{\sqrt{l^2 + (z - b)^2}} - \frac{1}{\sqrt{l^2 + (z - a)^2}}\right). \]
Misalkan $b = L \to \infty$ dan $a = -L \to -\infty$, maka
\[ E_z = \frac{\lambda}{4\pi\epsilon_0}\lim_{L\to\infty}\left(\frac{1}{\sqrt{l^2 + (z - L)^2}} - \frac{1}{\sqrt{l^2 + (z + L)^2}}\right) = 0. \]
\[ E_l := E_x\sec\phi = E_y\csc\phi = \frac{\lambda}{4\pi\epsilon_0l}\lim_{L\to\infty}\left(\frac{z + L}{\sqrt{l^2 + (z + L)^2}} - \frac{z - L}{\sqrt{l^2 + (z - L)^2}}\right). \]
\[ E_l = \frac{\lambda}{2\pi\epsilon_0l} \]
sesuai yang diharapkan.
\[ L(a, b) := \{(0, 0, z') ~|~ a < z' < b\} \]
di mana $a, b \in \mathbb{R}$ dan $a < b$.
Posisi titik pada $L(a, b)$ adalah $\vec{r}' := z'\hat{z}$ di mana $\hat{z} := (0, 0, 1)$.
Posisi sebarang titik pada ruang $\mathbb{R}^3$ adalah
\[ \vec{r} := \hat{x}l\cos\phi + \hat{y}l\sin\phi + \hat{z}z \]
di mana $\hat{x} := (1, 0, 0)$, $\hat{y} := (0, 1, 0)$, $l \in \mathbb{R}^+$, $\phi \in \{0\}\cup(0, 2\pi)$, dan $z \in \mathbb{R}$.
Medan listrik yang terjadi di titik $\vec{r}$ tentu saja adalah
\[ \vec{E} = \frac{\lambda}{4\pi\epsilon_0}\int_a^b \frac{\vec{r} - \vec{r}'}{|\vec{r} - \vec{r}'|^3}dz' \]
di mana $\epsilon_0$ merupakan permitivitas listrik dalam ruang hampa.
Tentu saja,
\[ \vec{E} = \frac{\lambda}{4\pi\epsilon_0}\int_a^b \frac{\hat{x}l\cos\phi + \hat{y}l\sin\phi + \hat{z}(z - z')}{[l^2 + (z - z')^2]^{3/2}}dz' = E_x\hat{x} + E_y\hat{y} + E_z\hat{z}. \]
Tentu saja,
\[ E_x = \frac{\lambda}{4\pi\epsilon_0}(l\cos\phi)\int_a^b \frac{dz'}{[l^2 + (z - z')^2]^{3/2}}. \]
Dengan substitusi $z - z' = l\tan\alpha$, maka diperoleh $-dz' = l\sec^2\alpha\,d\alpha$. Nilai $\alpha_a$ dan $\alpha_b$ didefinisikan sedemikian $z - a = l\tan\alpha_a$ dan $z - b = l\tan\alpha_b$, sehingga
\[ E_x = \frac{\lambda}{4\pi\epsilon_0}(l\cos\phi)\int_{\alpha_a}^{\alpha_b} \frac{-l\sec^2\alpha\,d\alpha}{l^3\sec^3\alpha}. \]
\[ E_x = -\frac{\lambda}{4\pi\epsilon_0l}(\cos\phi)\int_{\alpha_a}^{\alpha_b} \cos\alpha\,d\alpha. \]
\[ E_x = \frac{\lambda}{4\pi\epsilon_0l}(\cos\phi)(\sin\alpha_a - \sin\alpha_b). \]
\[ E_x = \frac{\lambda}{4\pi\epsilon_0l}\left(\frac{z - a}{\sqrt{l^2 + (z - a)^2}} - \frac{z - b}{\sqrt{l^2 + (z - b)^2}}\right)\cos\phi. \]
Dengan penalaran yang sama, diperoleh
\[ E_y = \frac{\lambda}{4\pi\epsilon_0l}\left(\frac{z - a}{\sqrt{l^2 + (z - a)^2}} - \frac{z - b}{\sqrt{l^2 + (z - b)^2}}\right)\sin\phi. \]
Komponen yang lain adalah
\[ E_z = \frac{\lambda}{4\pi\epsilon_0}\int_a^b \frac{z - z'}{[l^2 + (z - z')^2]^{3/2}}dz'. \]
\[ E_z = \frac{\lambda}{4\pi\epsilon_0}\int_{\alpha_a}^{\alpha_b} \frac{l\tan\alpha}{l^3\sec^3\alpha}(-1)l\sec^2\alpha\,d\alpha. \]
\[ E_z = -\frac{\lambda}{4\pi\epsilon_0l}\int_{\alpha_a}^{\alpha_b} \sin\alpha\,d\alpha. \]
\[ E_z = \frac{\lambda}{4\pi\epsilon_0l}(\cos\alpha_b - \cos\alpha_a). \]
\[ E_z = \frac{\lambda}{4\pi\epsilon_0l}\left(\frac{l}{\sqrt{l^2 + (z - b)^2}} - \frac{l}{\sqrt{l^2 + (z - a)^2}}\right). \]
\[ E_z = \frac{\lambda}{4\pi\epsilon_0}\left(\frac{1}{\sqrt{l^2 + (z - b)^2}} - \frac{1}{\sqrt{l^2 + (z - a)^2}}\right). \]
Misalkan $b = L \to \infty$ dan $a = -L \to -\infty$, maka
\[ E_z = \frac{\lambda}{4\pi\epsilon_0}\lim_{L\to\infty}\left(\frac{1}{\sqrt{l^2 + (z - L)^2}} - \frac{1}{\sqrt{l^2 + (z + L)^2}}\right) = 0. \]
\[ E_l := E_x\sec\phi = E_y\csc\phi = \frac{\lambda}{4\pi\epsilon_0l}\lim_{L\to\infty}\left(\frac{z + L}{\sqrt{l^2 + (z + L)^2}} - \frac{z - L}{\sqrt{l^2 + (z - L)^2}}\right). \]
\[ E_l = \frac{\lambda}{2\pi\epsilon_0l} \]
sesuai yang diharapkan.
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