Question

Answer the following questions:
(i) Define electric flux. Write its $SI$ units.
(ii) Using Gauss's law, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
(iii) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?

Answer 1

(a)

Electric flux is the measure of flow of the electric field through a given area. It is defined as the dot product of electric field and area vector through which the flux has to be computed. Hence,

$\varphi =\int \overrightarrow{E}.\overrightarrow{dS}$

Its SI unit is Gauss ($G$) / $N{m}^2{C}^{-1}$ / $Vm$.

(b)

Electric field near the surface of an infinite sheet of positive charge is directed away from the sheet of charge.

Consider a cylindrical gaussian surface with its axis perpendicular to the sheet of charge as shown in the figure.

The net flux through the curved surface is zero as the electric field and surface are parallel to each other.

Hence, ${\varphi }_{curved}=\int \overrightarrow{E}.\overrightarrow{dS}=0$

The net flux through the upper plane surface is given by:

${\varphi }_{pu}=\int \overrightarrow{E}.\overrightarrow{dS}=EA$

Similarly, The net flux through the lower plane surface is given by:

${\varphi }_{lu}=\int \overrightarrow{E}.\overrightarrow{dS}=EA$

Net flux through the closed cylindrical surface is:

$\varphi ={\varphi }_{curved}+{\varphi }_{pu}+{\varphi }_{lu}=2EA..........\left(i\right)$

By Gauss' Law,

$\varphi =\frac{Q_{enclosed}}{{\epsilon }_o}=\frac{\sigma A}{{\epsilon }_o}..............\left(ii\right)$

From (i) and (ii),

$E=\frac{\sigma }{2{\epsilon }_o}$

Hence, the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.

(c)

(i) If the sheet is positively charged, field is directed away from the sheet of charge.

(ii) If the sheet is negatively charged, field is directed towards the sheet of charge.