Mumbai University > Electronics Engineering > Sem 5 > Electromagnetic Engineering

Marks: 10 Marks

Year: May 2015

Consider an infinite surface charge with density $ρ_s$

According to gauss law:

$Q_{enclosed}=ρ_s .A$

For the four vertical sides of the box, $\bar{D} .d \bar{s}=0$

For top and bottom side, $\bar{D} .d \bar{s}=D_z.ds$

The flux through Gaussian surface is:

$ψ= ∮ \bar{D} .d \bar{s} \\ ψ= ∫_{top} \bar{D} .d \bar{s}+ ∫_{bottom} \bar{D} .d \bar{s}+ ∫_{left} \bar{D} .d \bar{s }+ ∫_{right} \bar{D} .d \bar{s}+ ∫_{front} \bar{D} .d \bar{s}+ ∫_{back} \bar{D} .d \bar{s} \\ ψ= ∫_{top} \bar{D} .d \bar{s}+ ∫_{bottom} \bar{D} .d \bar{s} \\ ψ=D_z ∫_{top}ds+ D_z ∫_{bottom}ds \\ ψ=D_z A+D_z A \\ ψ=2D_z.A$

Total flux crossing the surface = charges enclosed

$2A.D_z=ρ_s.A \\ D_z=\dfrac{ρ_s}{2} \\ \bar{D }=D_z \overline{a_z} =\dfrac{ρ_s}{2} \overline{a_z} (c/m^2)$

The field intensity is

$\bar{E}=\dfrac{\bar{D}}{ε_o} \\ \bar{E}=\dfrac{ρ_s}{2ε_o} \overline{a_z}$