Question

Find the electric field at a distance "$z$" from the plane.

• A. $\overrightarrow{E}=\frac{\sigma }{2{ϵ}_0}\hat{k}$ for $z>0$
• B. $\overrightarrow{E}=-\frac{\sigma }{2{ϵ}_0}\hat{k}$ for $z>0$
• C. $\overrightarrow{E}=\frac{\sigma }{2{ϵ}_0}\hat{k}$ for $z<0$
• D. $\overrightarrow{E}=-\frac{\sigma }{2{ϵ}_0}\hat{k}$ for $z<0$

Answer 1

Answer: (A), (D)

The correct options are
A $\overrightarrow{E}=\frac{\sigma }{2{ϵ}_0}\hat{k}$ for $z>0$
D $\overrightarrow{E}=-\frac{\sigma }{2{ϵ}_0}\hat{k}$ for $z<0$

Consider an infinite large non-conducting plane in the x-y plane with uniform charge density $\sigma$.

To calculate the electric field $\overrightarrow{E}$ at a point P at distance "$x$" from the plane, draw a gaussian surface in the form of a closed cylinder of length $x$ on each side of the plane with end caps of area $S$. Electric field $\overrightarrow{E}$is perpendicular to the plane and hence it is normal to the end caps $I$ and $II$ also.

Charge enclosed by gaussian surface $q=\sigma S$

According to Gauss Theorem,

$\oint \overrightarrow{E}.\overrightarrow{dS}=\frac{q}{{\epsilon }_0}=\frac{\sigma S}{{\epsilon }_0}$ ...(i)

Gussian surface is divided into three parts $I,II,III$ i.e, end caps and curved surface of the cylinder as shown in the figure.

$\therefore$ equation (i) becomes:

${\oint }_I\overrightarrow{E}.\overrightarrow{dS}+{\oint }_{II}\overrightarrow{E}.\overrightarrow{dS}+{\oint }_{III}\overrightarrow{E}.\overrightarrow{dS}=\frac{\sigma S}{{\epsilon }_0}$ ...(ii)

Angle between $\overrightarrow{E}$ and $\overrightarrow{dS}$ is ${90}^{\circ }$ for curved surface $III$. In this case, ${\oint }_{III}\overrightarrow{E}.\overrightarrow{dS}=EdS\cos {90}^{\circ }=0$

Hence, equation (ii) becomes ${\oint }_I\overrightarrow{E}.\overrightarrow{dS}+{\oint }_{II}\overrightarrow{E}.\overrightarrow{dS}=\frac{\sigma S}{{\epsilon }_0}$

Since angle between $\overrightarrow{E}$ and $\overrightarrow{dS}$ for surfaces $I$ and $II$ is zero, so:

${\oint }_{I}EdS\cos 0^{\circ }+{\oint }_{II}EdS\cos 0^{\circ }=\frac{\sigma S}{{\epsilon }_0}$

or, ${\oint }_{I}EdS+{\oint }_{II}EdS=\frac{\sigma S}{{\epsilon }_0}$

Since electric field intensity $E$ is constant at every point of the gaussian surface so,

$E{\oint }_{I}ds+E{\oint }_{II}ds=\frac{\sigma S}{{\epsilon }_0}$

or, $ES+ES=\frac{\sigma S}{{\epsilon }_0}$

or, $2ES=\frac{\sigma S}{{\epsilon }_0}$

or, $E=\frac{\sigma }{2{\epsilon }_0}$

In vector form, $\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_0}\hat{n}$ where $\hat{n}=$ unti vector perpendicular to the plane of sheet pointing away from it i.e, $\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_0}\hat{k}$

The electric field is directed away from plane if it is positively charged i.e,

$\overrightarrow{E}=\frac{\sigma }{2{\epsilon }_0}\hat{k}(z>0)$

and it is diercted towards the plane if it is negatively charged i.e,

$\overrightarrow{E}=\frac{-\sigma }{2{\epsilon }_0}\hat{k}(z<0)$