## Electric Field Due to a Long Charge Carrying Wire

We can find the electric field around a long, cuharge carring wire, carrying charge
$\beta$
per unit length. For a surface
$S$
enclosing a charge
$q$
, Gauss's Law gives
$\int_S \mathbf{E} \cdot \mathbf{n} dS = \frac{q}{\epsilon_0}$

The cylinder drawn has three surfaces. For surfacesat the ends of the cylinder the electric field, which is radial, is perpendicular to the normal, which is along the wire, so
$\int_{S_{TOP}} \mathbf{E} \cdot \mathbf{n} dS = \int_{S_{BOTTOM}} \mathbf{E} \cdot \mathbf{n} dS =0$
and only the curved surface of the cylinder contributues to the integral.
For the curved surface
$\mathbf{E}$
is radially out by symmetry and the normal is also radially out. Hence, using cylindrical polar coordinates,
\begin{equation} \begin{aligned} \int_S \mathbf{E} \cdot \mathbf{n} dS &= \int^L_0 \int^{2 \pi}_0 E r d \theta dz \\ &= 2 \pi r L E \end{aligned} \end{equation}

Hence
$2 \pi r LE = \frac{L \beta}{\epsilon_0} \rightarrow \mathbf{E} = \frac{\beta}{2 \pi r} \mathbf{e_r}$ 