JEE Main & Advanced Physics Electrostatics & Capacitance Gauss's Law and it's Application

(1) According to this law, the total flux linked with a closed surface called Gaussian surface. (The surface need not be a real physical surface, it can also be an hypothetical one) is \[(1/{{\varepsilon }_{0}})\] times the charge enclosed by the closed surface i.e., \[\varphi =\oint_{s}{\,\overrightarrow{E}\cdot \overrightarrow{dA}=\frac{1}{{{\varepsilon }_{o}}}({{Q}_{enc}})}\]

(2) Electric field in \[\oint{\overrightarrow{E}.\,\overrightarrow{dA}}\] is complete electric field. It may be partly due to charge with in the surface and partly due to charge outside the surface. However if there is no charge enclosed in the Gaussian surface, then \[\oint{\overrightarrow{E}.\,\overrightarrow{dA}}=0\].

(3) The electric field \[\overrightarrow{E}\] is resulting from all charge, both those inside and those outside the Gaussian surface.

(Keep in mind, the electric field due to a charge outside the Gaussian surface contributes zero net flux through the surface, Because as many lines due to that charge enter the surface as leave it).

Flux from surface \[{{S}_{1}}=+\,\frac{Q}{{{\varepsilon }_{0}}},\] Flux from surface \[{{S}_{2}}=-\,\frac{Q}{{{\varepsilon }_{0}}},\] and flux from \[{{S}_{3}}=\] flux from surface \[{{S}_{4}}=0\]

Application of Gauss's law : See flux emergence in the following cases

(1) If a dipole is enclosed by a surface

\[\because \,{{Q}_{enc}}=0\]

\[\Rightarrow \]\[\varphi =0\]

(2) The net charge \[{{Q}_{enc}}\] is the algebraic sum of all the enclosed positive, and negative charges. If \[{{Q}_{enc}}\] is positive the net flux is outward; if \[{{Q}_{enc}}\] is negative, the net flux is inward.

\[\varphi =\frac{1}{{{\varepsilon }_{0}}}({{Q}_{1}}+{{Q}_{2}}-{{Q}_{3}})\]

(3) If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non-uniform) total flux linked with it will be zero

(4) If a hemispherical body is placed in uniform electric field then flux linked with the curved surface calculated as follows

\[{{\varphi }_{Curved}}+{{\varphi }_{Circular}}=0\]

\[{{\varphi }_{Curved}}=-{{\varphi }_{Circular}}\]

\[=-\,(E\times \pi {{R}^{2}}\cos 180{}^\circ )\]

\[=+\pi {{R}^{2}}E\]

(5) If a hemispherical body is placed in non-uniform electric field as shown below. then flux linked with the circular surface calculated as follows

\[{{\varphi }_{Circular}}=-{{\varphi }_{Curved}}\]

\[{{\varphi }_{Circular}}=-\,(E\times 2\pi {{R}^{2}}\cos 0{}^\circ )\]

\[=-\,2\pi {{R}^{2}}E\]

(6) If charge is kept at the centre of cube

\[{{\varphi }_{total}}=\frac{1}{{{\varepsilon }_{0}}}.(Q)\]   

\[{{\varphi }_{face}}=\frac{Q}{6{{\varepsilon }_{0}}}\]

\[{{\varphi }_{corner}}=\frac{Q}{8{{\varepsilon }_{0}}}\]              

\[{{\varphi }_{edge}}=\frac{Q}{12{{\varepsilon }_{0}}}\]

(7) If charge is kept at the centre of a face : First we should enclosed the charge by assuming a Gaussian surface (an identical imaginary cube)

Total flux emerges from the system (Two cubes) \[{{\varphi }_{total}}=\frac{Q}{{{\varepsilon }_{0}}}\]

Flux from given cube (i.e. from 5 face only) \[{{\varphi }_{cube}}=\frac{Q}{2{{\varepsilon }_{0}}}\]

(8) If a charge is kept at the corner of a cube

For enclosing the charge seven more cubes are required so total flux from the 8 cube system is \[{{\phi }_{1}}=\frac{Q}{{{\varepsilon }_{0}}}\]. Flux from given cube \[{{\varphi }_{cube}}=\frac{Q}{8{{\varepsilon }_{0}}}\]. Flux from one face opposite to change, of the given cube \[{{\varphi }_{face}}=\frac{Q/8{{\varepsilon }_{0}}}{3}=\frac{Q}{24{{\varepsilon }_{0}}}\] (Because only three faces are seen).

(9) A long straight wire of charge density \[\lambda \] penetrates a hollow body as shown. The flux emerges from the body is

\[\phi =\lambda \times \] (Length of the wire inside the body)