Question
Get the electric field due to a uniformly charged infinite plane sheet.
✓
Answer
►Let $\sigma$ be the uniform surface charge density of an infinite plane sheet (Fig.). We take the $x$-axis normal to the given plane.
►By symmetry, the electric field will not depend on y and z coordinates and its direction at every point must be parallel to the x-direction.
►We can take the Gaussian surface to be a rectangular parallelpiped of cross-sectional area A, as shown. (A cylindrical surface will also do.)
►As seen from the figure, only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel to the other faces and they, therefore, do not contribute to the total flux.
►Total flux passing through Gaussian surface.
$\begin{aligned}
\phi & =\text { Electric flux passing through Surface } 1 \\
& + \text { Electric flux passing through surface } 2 \\
\therefore \phi & =\text { EAcos } 0+\text { EA } \cos 0 \\
& =\text { EA }+ \text { EA } \\
& =2 EA
\end{aligned}$
►According to Gauss's law $=\phi=\frac{q}{\varepsilon_0}$
Thus, 2EA $=\frac{q}{\varepsilon_0}$
Where, $q=$ electric charge enclosed by
Gaussian surface
$\begin{aligned}
\quad q & =\text { Surface charge density } \times \text { Area } \\
\therefore \quad q & =\sigma A
\end{aligned}$
►Put the value in equation (3),
$\begin{array}{rlrl}
\therefore & 2 EA & =\frac{\sigma A }{\varepsilon_0} \\
\therefore & E & =\frac{\sigma}{2 \varepsilon_0} \\
2 EA & =\frac{\sigma A }{\varepsilon_0} \\
& \text { or, } & E & =\frac{\sigma}{2 \varepsilon_0}
\end{array}$
►Vectorically,
$\overrightarrow{ E }=\frac{\sigma}{2 \varepsilon_0} \hat{n}$
►Where $\hat{n}$ is a unit vector normal to the plane and going away from it, E is directed away from the plate if $\sigma$ is positive and toward the plate if $\sigma$ is negative.
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