• # question_answer State Gauss's law in electrostatics. Using this law, derive an expression for the electric field due to a uniformly charged infinite plane sheet.

Gauss's law states that the total electric flux linked with a closed surface is equal to $1/{{\varepsilon }_{0}}$ times total charge, enclosed by that closed (Gaussian) surface, i.e. ${{\phi }_{E}}=\oint{E\cdot dS=\frac{{{q}_{in}}}{{{\varepsilon }_{0}}}}$ where, ${{q}_{in}}$= total charge inside the closed (Gaussian) surface Let us consider, a large plane sheet of charge having surface charge density $\sigma .$ Let electric field is to be obtained at a point P at a distance r from it. It is obvious that Gaussian surface will be a cylinder of cross-sectional area A and length 2r with its axis perpendicular to plane sheet of charge. Now, applying Gauss's law over the closed Gaussian surface,             $\oint_{S}{E\cdot dS}=\frac{q}{{{\varepsilon }_{0}}}\Rightarrow \int_{caps}{E.dS}+\int_{CSA}{E\cdot dS}=\frac{q}{{{\varepsilon }_{0}}}$ Note Closed cylinder comprises of two caps and Curved Surface Area (CSA). $\int_{caps}{E\,dS\cos 0{}^\circ +\int_{CSA}{E\,dS}\cos 90{}^\circ =\frac{q}{{{\varepsilon }_{0}}}}$ (At caps E and dS are along the same direction whereas $CSA\,E\bot dS$) $\therefore$      $E\int_{caps}{dS=\frac{q}{{{\varepsilon }_{0}}}\Rightarrow E\times 2A=\frac{q}{{{\varepsilon }_{0}}}}$ $\Rightarrow$   $E=\frac{q}{2A{{\varepsilon }_{0}}}=\frac{\sigma }{2{{\varepsilon }_{0}}}$     $\left[ \because \sigma =\frac{q}{A} \right]$ The direction of E is normal to plane sheet and directed away from sheet (when charge on plate is positive and vice-versa).