Question

Write the statement of Gauss' law for electrostatics.
Draw a diagram and derive an expression for electric field due to a uniformly charged infinite plane sheet at a point near the sheet.
For the given diagram, write the value of electric flux passing through the surface.

Answer 1

$\text{Step 1: Statement of Gauss's law}$

Gauss Law for electrostatics states that the total electric flux passing through a closed surface equals the enclosed charge in the surface divided by the permittivity of the medium.

${\varphi }_E=\oint (\overrightarrow{E}\cdot \overrightarrow{dS})=\frac{Q_{enclosed}}{{\epsilon }_o}$

$\text{Step 2: Deriving an expression for electric field due to infinite plane sheet}$

Consider an infinite plane sheet of positive charge with charge density $\sigma$ as shown in the attached figure. Electric field lines will be directed orthogonal and away from the sheet of charge. Hence, a cylindrical closed surface with its base parallel to the sheet of paper as shown in the figure is a good choice of Gaussian surface.

For the curved surface electric field is orthogonal to the surface area vector. Hence, flux linked to curved surface is zero.

${\varphi }_s=0$

For the plane surface, applying Gauss's Law, we get:

${\varphi }_s+{\varphi }_b=\frac{Q_{enclosed}}{{\epsilon }_o}$

$EA+EA=\frac{Q_{enclosed}}{{\epsilon }_o}$

But $Q=\sigma A$ by definition of surface charge density

$⟹E=\frac{\sigma }{2{\epsilon }_o}$

[Note : For an infinite line sheet of charge, electric filed does not depend on the distance of the point from the surface.]

$\text{Step 3: Electric flux passing through surface in given diagram}$

For the diagram shown in the question, enclosed charge is:

$Q=2+(-1)=1\mu C$

Using Gauss's Law,

Electric flux through the surface is:

${\varphi }_E=\frac{Q}{{\epsilon }_o}$

$={10}^{-6}\times 4\pi \times 9\times {10}^9$

$\approx 1.13\times {10}^{5}Vm$