Question

A large plane charged sheet having surface charge density $\sigma =+2\times {10}^{-6}{\text{C/m}}^2$ lies in the $\text{x-y}$ plane. Find the flux of the electric field through a circular area of radius $1\text{cm}$ lying completely in the region where $\text{x, y, z}$ all are positive and with its normal making an angle of ${60}^{\circ }$ with the $\text{z-}$axis.

• A. $17.3{\text{Nm}}^2\text{/C}$
• B. $16.3{\text{Nm}}^2\text{/C}$
• C. $8.15{\text{Nm}}^2\text{/C}$
• D. $34.6{\text{Nm}}^2\text{/C}$

Answer 1

The correct option is A $17.3{\text{Nm}}^2\text{/C}$

Given,
Surface charge density of the given sheet, $\sigma =2\times {10}^{-6}{\text{C/m}}^2$
Radius of circular area, $r=1\text{cm}$


Electric field near the plane charged sheet,
$E=\frac{\sigma }{2{ϵ}_0}$
Substituting the values ,
$E=\frac{2\times {10}^{-6}}{2\times 8.85\times {10}^{-12}}=1.1\times {10}^5\text{N/C}$

And the electric field acts in the $\text{z-}$ direction (normal to the sheet).

Area of given surface,
$A=\pi r^2$
$\Rightarrow A=3.14\times (0.01{)}^2=3.14\times {10}^{-4}{\text{m}}^2$

Positive normal to it make an angle of ${60}^{\circ }$ with $\text{z-}$axis.
$\therefore \theta ={60}^{\circ }$

Electric flux through the given surface,
$\varphi =\overrightarrow{E}.\overrightarrow{A}=EA\cos \theta$
Substituting the values,
$\varphi =1.1\times {10}^5\times 3.14\times {10}^{-4}\times \cos {60}^{\circ }$
$\therefore \varphi =17.3{\text{N-m}}^2\text{/C}$