The electric field near a conducting surface | vedclass

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Question

(C) According to Gauss's Law,for a charged conductor,the electric field just outside the surface is perpendicular to the surface.By applying Gauss's L

Vedclass · Accepted Answer

$\frac{\sigma}{\varepsilon_0}$ and is normal to the surface

Vedclass · Answer

(C) According to Gauss's Law,for a charged conductor,the electric field just outside the surface is perpendicular to the surface.By applying Gauss's Law to a small cylindrical Gaussian surface enclosing a charge $\sigma A$ on the conductor,we have $\oint E \cdot dA = \frac{q_{enclosed}}{\varepsilon_0}$.Since the electric field inside the conductor is zero and the field is perpendicular to the surface,the flux through the curved surface is zero,and the flux through the inner flat face is zero.Thus,$E \cdot A = \frac{\sigma A}{\varepsilon_0}$,which simplifies to $E = \frac{\sigma}{\varepsilon_0}$.Therefore,the electric field is $\frac{\sigma}{\varepsilon_0}$ and is directed normal to the surface.

The electric field near a conducting surface having a uniform surface charge density $\sigma$ is given by:

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