Question
State Gauss theorem in electrostatics. Apply this theorem to obtain the expression for the electric field at a point due to an infinitely long, thin, uniformly charged straight wire of linear charge density $\lambda\text{Cm}^{-1}.$
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Answer
Gauss Theorem: It states that total electric flux over the closed surface S is $\frac{1}{\in_0}$ times the total charge (q) contained in side S.$\therefore\ \Phi_\text{E}=\oint\limits_{\text{S}}\vec{\text{E}}.\vec{\text{dS}}=\frac{\text{q}}{\in_0}$
Electric field due to infinitely long, thin and uniformly charged straight wire: Consider an infinitely long line charge having linear charge density $\lambda$ coulomb metre$^{-1}$ (linear charge density means charge per unit length). To find the electric field strength at a distance r, we consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts:
We consider small elements of surfaces $S_1, S_2$ and $S_3$ The surface element vector $\vec{\text{dS}}_1$ is directed along the direction of electric field (i.e., angle between $\vec{\text{E}}$ and $\vec{\text{dS}}_1$ is zero); the elements $\vec{\text{dS}}_2$ and $\vec{\text{dS}}_3$ are directed perpendicular to field vector $\vec{\text{E}}$ (i.e., angle between $\vec{\text{dS}}_2$ and $\vec{\text{E}}$ is 90° and so also angle between $\vec{\text{dS}}_3$ and $\vec{\text{E}}).$
Electric Flux through the cylindrical surface,
Image
$\oint\limits_{\text{S}}\vec{\text{E}}.\vec{\text{dS}}=\oint\limits_{\text{S}_1}\vec{\text{E}}.\vec{\text{dS}_1}+\oint\limits_{\text{S}_2}\vec{\text{E}}.\vec{\text{dS}_2}+\oint\limits_{\text{S}_3}\vec{\text{E}}.\vec{\text{dS}_3}$
$=\oint\limits_{\text{S}_1}{\text{E}}.{\text{dS}_1}\cos0^\circ+\oint\limits_{\text{S}_2}{\text{E}}.{\text{dS}_2}\cos90^\circ+\oint\limits_{\text{S}_3}{\text{E}}.{\text{dS}_3}\cos90^\circ$
$=\oint\text{E dS}_1+0+0$
$=\text{E}\oint\text{dS}_1$ (since electric field E is the same at each point of curved surface)
$=\text{E}2\pi\text{rl}$ (since area of curved surface $=2\pi\text{rl})$
As $\lambda$ is charge per unit length and length of cylinder is l therefore, charge enclosed by assumed surface $=(\lambda\text{l})$
$\therefore$ By Gauss's theorem
$\int\vec{\text{E}}.\vec{\text{dS}}=\frac{1}{\in_0}\times\text{charge enclosed}$
$\Rightarrow\ \text{E}.2\pi\text{rl}=\frac{1}{\in_0}(\lambda\text{l})$
$\Rightarrow\ \text{E}=\frac{\lambda}{2\pi\in_0\text{r}}$
Thus, the electric field strength due to a line charge is inversely proportional to r.
Electric field due to infinitely long, thin and uniformly charged straight wire: Consider an infinitely long line charge having linear charge density $\lambda$ coulomb metre$^{-1}$ (linear charge density means charge per unit length). To find the electric field strength at a distance r, we consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts:
- Curved surface $S_1.$
- Flat surface $S_2.$
- Flat surface $S_3.$
We consider small elements of surfaces $S_1, S_2$ and $S_3$ The surface element vector $\vec{\text{dS}}_1$ is directed along the direction of electric field (i.e., angle between $\vec{\text{E}}$ and $\vec{\text{dS}}_1$ is zero); the elements $\vec{\text{dS}}_2$ and $\vec{\text{dS}}_3$ are directed perpendicular to field vector $\vec{\text{E}}$ (i.e., angle between $\vec{\text{dS}}_2$ and $\vec{\text{E}}$ is 90° and so also angle between $\vec{\text{dS}}_3$ and $\vec{\text{E}}).$
Electric Flux through the cylindrical surface,
Image
$\oint\limits_{\text{S}}\vec{\text{E}}.\vec{\text{dS}}=\oint\limits_{\text{S}_1}\vec{\text{E}}.\vec{\text{dS}_1}+\oint\limits_{\text{S}_2}\vec{\text{E}}.\vec{\text{dS}_2}+\oint\limits_{\text{S}_3}\vec{\text{E}}.\vec{\text{dS}_3}$
$=\oint\limits_{\text{S}_1}{\text{E}}.{\text{dS}_1}\cos0^\circ+\oint\limits_{\text{S}_2}{\text{E}}.{\text{dS}_2}\cos90^\circ+\oint\limits_{\text{S}_3}{\text{E}}.{\text{dS}_3}\cos90^\circ$
$=\oint\text{E dS}_1+0+0$
$=\text{E}\oint\text{dS}_1$ (since electric field E is the same at each point of curved surface)
$=\text{E}2\pi\text{rl}$ (since area of curved surface $=2\pi\text{rl})$
As $\lambda$ is charge per unit length and length of cylinder is l therefore, charge enclosed by assumed surface $=(\lambda\text{l})$
$\therefore$ By Gauss's theorem
$\int\vec{\text{E}}.\vec{\text{dS}}=\frac{1}{\in_0}\times\text{charge enclosed}$
$\Rightarrow\ \text{E}.2\pi\text{rl}=\frac{1}{\in_0}(\lambda\text{l})$
$\Rightarrow\ \text{E}=\frac{\lambda}{2\pi\in_0\text{r}}$
Thus, the electric field strength due to a line charge is inversely proportional to r.
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