Name: Electric Field of a Charged Plane Sheet
Uploaded: 2019-07-19
Duration: 6 min 58 s
Description: Another distribution of charge. This time on an infinite plane sheet. Let’s see once again how Gauss’s law comes to the rescue in electric field calculations.

Question

State and prove Gauss's Law in Electrostatics.

Accepted Answer

This theorem gives a relationship between the total flux passing through any closed surface and the net charge enclosed within the surface.

Gauss theorem states that the total flux through a closed surface is $\frac{1}{ε_{0}}$ times the net charge enclosed by the closed surface.

Mathematically, it can be expressed as

$ϕ_{E} = \oint \vec{E} \cdot d \vec{S} = \frac{q}{ε_{0}}$ ..............................................(1)

Where $\vec{E}$ is the electric field and $S$ = surface area , $ε_{0}$ = permittivity of free space.

Proof:

Consider an isolated positive point charge $q$ , , suppose the surface $S$ is a sphere of radius $r$ with $q$ as centre.

Electric field at any point on $S$ is

$E = \frac{1}{4 {πε}_{0}} \cdot \frac{q}{r^{2}}$ ..........................................(2)

This field points radially outward at all points on $S$ Also, any area element points radially outwards, so it is parallel to $\vec{E}$ , i.e., angle between $\vec{E}$ and area element, $θ = 0 °$ .

$∴$ Flux through area $d \vec{S}$ is ${dϕ}_{E} = \vec{E} \cdot d \vec{S} = E dS \cos 0^{\circ} = EdS$ ,

Total flux through surface $S$ is,

$ϕ_{E} = \oint_{S} {dϕ}_{E} = \oint_{S} EdS = E \oint_{S} dS$ $= E \times total area of sphere$ .

$= \frac{1}{4 {πε}_{0}} \cdot \frac{q}{r^{2}} \times 4 {πr}^{2}$ .

$ϕ_{E} = \frac{q}{ε_{0}}$ .

Hence proved.

State and prove Gauss's Law in Electrostatics.

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