Question

Electric field due to an infinite sheet of charge having surface density $\sigma$ is $E$. The electric field due to an infinite conducting sheet of the same surface density of charge is
A. ${\displaystyle \frac{E}{2}}$
B. $E$
C. $2E$
D. $4E$

Answer 1

Hint: The electric field of the infinite charged sheet can be calculated using the Gauss theorem. A cylindrical Gaussian surface is considered, which is intersecting the sheet. By using the Gauss law, the electric field on the three surfaces is derived.

Formula used:
Gauss law states that,
$\varphi ={\displaystyle \frac{q}{{\epsilon }_0}}$
Where $\varphi$ is the electric flux, $q$ is the charge, and ${\epsilon }_0$ is the electric constant.

Complete step by step solution:

Consider a Gaussian volume as a cylinder, which intersects as shown in figure. The total flux over the Gaussian surface is defined by,
Total flux = Flux in curved surface + Flux in flat surface 1 + Flux in flat surface 2
$\varphi =\oint {\overrightarrow{E}}_1.d\overrightarrow{s}=\underset{Curve}{\int }{E}_1.ds.\cos \theta +\underset{Flat-1}{\int }{E}_1.ds.\cos \theta +\underset{Flat-2}{\int }{E}_1.ds.\cos \theta$
Since, the angle between electric field of lines and curved face, flat face-1 and flat face-2 are ${90}^{\circ }$, $0^{\circ }$ and $0^{\circ }$.
Thus,
$\varphi =\underset{Curve}{\int }{E}_1.ds.\cos 90+\underset{Flat-1}{\int }{E}_1.ds.\cos 0+\underset{Flat-2}{\int }{E}_1.ds.\cos 0$
Since, $\cos 90=0$ and $\cos 0=1$
Then,
$$\varphi =0+\underset{}{\int }{E}_1.ds+\underset{}{\int }{E}_1.ds$$
Since, $E$ is always constant.
Then,
$$\varphi =E_1\underset{}{\int }ds+E_1\underset{}{\int }ds=2E_1\int ds$$
The value of the integral $\int ds$ is area. So, $\int ds=A$
Hence, $\varphi =2E_{1}A.............................................\left(1\right)$
By Gauss theorem,
$\varphi ={\displaystyle \frac{q}{{\epsilon }_0}}.......................................\left(2\right)$
By equating the equations (1) and (2), we get
$$\begin{gathered} 2E_{1}A={\displaystyle \frac{q}{{\epsilon }_0}} \\ {E}_1={\displaystyle \frac{q}{2{\epsilon }_{0}A}} \end{gathered}$$
The charge inside the Gaussian surface is defined by, $q=\sigma A$
Where $\sigma$ is the charge density and $A$ is the area.
Thus,
$$\begin{gathered} E_1={\displaystyle \frac{\sigma A}{2{\epsilon }_{0}A}} \\ {E}_1={\displaystyle \frac{\sigma }{2{\epsilon }_0}} \end{gathered}$$
Since, Electric field due to an infinite sheet of charge having surface density $\sigma$ is $E={\displaystyle \frac{\sigma }{{\epsilon }_0}}$
Then, $E_1={\displaystyle \frac{E}{2}}$

$\therefore$ Electric field due to an infinite conducting sheet of the same surface density of charge is ${\displaystyle \frac{E}{2}}$. Hence the option (A) is correct.

Note:
The sheet is a conducting sheet, so the electric field is half of the normal infinite sheet. The Gaussian surface must be intersected through the plane of the conducting sheet. The electric field is completely dependent on the charge density and the area of the surface and also depends on the electric constant.