## Electric Field Due to a Long Charge Carrying Wire

\[\beta\]

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

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