Electric Field Due to a Long Charge Carrying Wire

We can find the electric field around a long, cuharge carring wire, carrying charge  
  per unit length.

For a surface  
  enclosing a charge  
, 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  
  is radially out by symmetry and the normal is also radially out. Hence, using cylindrical polar coordinates,
\[\begin{equation} \begin{aligned} \int_S \mathbf{E} \cdot \mathbf{n} dS &= \int^L_0 \int^{2 \pi}_0 E r d \theta dz \\ &= 2 \pi r L E \end{aligned} \end{equation}\]

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

You have no rights to post comments