Obtain an expression for electric field at the surface of a charged conductor.
Obtain an expression for electric field at the surface of a charged conductor.
We consider a Gaussian surface of a pill box of extremely small length and extremely small cross-section as $\sigma$.
A fraction of it is inside the surface and the remaining part is outside the surface.
The total charge enclosed by this pill box is $q=\sigma d s$
where $\sigma=$ surface charge density of conductor.
At every point on the surface of the conductor $\vec{E}$ is perpendicular to the surface. Hence, it is parallel to the surface vector $\overrightarrow{\mathrm{E}} \| d \vec{s}$
But inside the surface $\vec{E}=0$. Hence, the flux coming out from the cross-section of pill box inside the surface $=0$.
The flux coming out from the cross-section of pill box outside the surface, $\phi=\overrightarrow{\mathrm{E}} \cdot d \vec{s}=\mathrm{E} d s \cos 0^{\circ}=\mathrm{E} d s$
According to Gauss's theorem,
$\phi=\mathrm{E} d s$
$\therefore \frac{q}{\varepsilon_{0}}=\mathrm{E} d s$
$\therefore \frac{\sigma d s}{\varepsilon_{0}}=\mathrm{E} d s$
$\therefore \mathrm{E}=\frac{\sigma}{\varepsilon_{0}}$
In the vector form $\overrightarrow{\mathrm{E}}=\frac{\sigma}{\varepsilon_{0}} \cdot \hat{n}$
If $\sigma$ is positive $\overrightarrow{\mathrm{E}}$ is in the direction of normal coming out from the surface. If $\sigma$ is negative $\overrightarrow{\mathrm{E}}$ is in direction of normal entering into the surface.
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