Question: Use Gauss's theorem to find electric field intensity due to an infinitely long straight Uniformly charged thin wire or conductor.
Solution:
Consider an infinite and very thin straight uniformly charged wire having linear charge density (i.e., charge per unit length) $\lambda$.
To calculate the electric field intensity $E$ at a point $P$, distant $r$ from the line, draw an imaginary cylinder (Gaussian surface) of radius $r$ and length $l$ around the charged line.
The charge enclosed by the Gaussian surface,
\[ q = \lambda l \]
According to Gauss' theorem,
\[ \oint_S \vec{E} \cdot d\vec{S} = \frac{q}{\epsilon_0} = \frac{\lambda l}{\epsilon_0} \qquad \text{...(1)} \]
The cylindrical Gaussian surface is divided into three parts I, II and III i.e. top circular face, bottom circular face and curved surface respectively as shown in figure 63.
Therefore, eqn. (1) can be written as,
$\oint \vec{E} \cdot d\vec{S} =$
$=\int_I \vec{E} \cdot d\vec{S} + \int_{II} \vec{E} \cdot d\vec{S} + \int_{III} \vec{E} \cdot d\vec{S}$
\[ \oint \vec{E} \cdot d\vec{S} = \frac{\lambda l}{\epsilon_0} \qquad \text{...(2)} \]
For surfaces I and II, angle between $\vec{E}$ and $d\vec{S}$ is $90^\circ$, so $\vec{E} \cdot d\vec{S} = E \, dS \, \cos \theta = E \, dS \, \cos 90^\circ = 0$ for these surfaces. Therefore, electric flux will cross through the curved surface only.
Therefore, eqn. (2) becomes
\[ \oint_S \vec{E} \cdot d\vec{S} = \int_{III} \vec{E} \cdot d\vec{S} = \frac{\lambda l}{\epsilon_0} \]
or
\[ \int_{III} \vec{E} \cdot d\vec{S} = \frac{\lambda l}{\epsilon_0} \]
or
\[ \int_{III} E \, dS \, \cos 0^\circ = \frac{\lambda l}{\epsilon_0} \]
or
\[ \int_{III} E \, dS = \frac{\lambda l}{\epsilon_0} \]
Since electric field intensity $E$ is constant at every point of the Gaussian surface, so,
\[ E \int_{III} dS = \frac{\lambda l}{\epsilon_0} \qquad \text{...(3)} \]
But $\int_{III} dS$ = area of the curved surface of the cylinder $= 2\pi rl$
\[ \therefore \quad E \times 2\pi rl = \frac{\lambda l}{\epsilon_0} \quad \text{or} \quad E = \frac{\lambda}{2\pi \epsilon_0 r} \qquad \text{...(4)} \]
Thus
\[ E \propto \frac{1}{r} \]
In vector form, eqn. (4) can be written as $\vec{E} = \frac{\lambda}{2\pi \epsilon_0 r} \hat{n}$, where $\hat{n}$ is a unit vector perpendicular to the curved surface of the wire.
The variation of electric field $E$ with distance $r$ perpendicular to the line of charge.
