Electric Field Due to Infinite Uniform Positive Charged Plate

We can use Gauss's Law to find the electric field due an an infinite flat plate with a uniform positive surface charge  
. We enclose a section of the plate within a closed cuboid and apply Gauss's Law to the surface.

Gauss's Law states  
\[\int_S \mathbf{E} \cdot \mathbf{n} dS = \frac{q}{\epsilon_0}\]

  is the charge enclosed by the surface. By summetry  
  is constant and perpendicular to the surface of the plate. This means that at the sides of the surface the normal  
  is at right angles to the electric field  
. The only contributions to the surface integral comes from the surfaces of the cuboid parallel to the plate.
\[\int_S \mathbf{E} \cdot \mathbf{n} dS = \frac{q}{\epsilon_0} \rightarrow 2 AE = \alpha A \rightarrow E =\frac{A \alpha}{\epsilon_0} \rightarrow E = \frac{\alpha}{2 \epsilon_0}\]

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