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}+ …