Question

A large plane charge sheet having surface charge density $\sigma =2.0\times {10}^{-6}C{m}^{-2}$ lies in the x-y plane. Find the flux of the electric field through a circular area of radius $1cm$ lying completely in the region where x, y, z are all positive and with its normal making an angle of ${60}^{\circ }$ with the z-axis. (Given that the field due to infinite charged sheet is $\frac{\sigma }{2{ϵ}_0}$ where $\sigma$ = surface charge density. Also note that it is perpendicular to the sheet and is constant for all points from the sheet.)

• A. $35\frac{N{m}^2}{C}$
• B. $35\frac{C}{N{m}^2}$
• C. $17.5\frac{N{m}^2}{C}$
• D. $17.5\frac{C}{N{m}^2}$

Answer 1

The correct option is C

$17.5\frac{N{m}^2}{C}$

Now it is given that field is constant but area vector makes some angle with the direction of $\overrightarrow{E}$ field. You clearly know that in such cases there is a simplified notation as.

$\varphi =\overrightarrow{E}.\overrightarrow{A}$

The normal to the x-y plane is the z axis and the electric field will be along the z axis. Now from the question, the angle between $\overrightarrow{E}$ and $\overrightarrow{A}$ is ${60}^0$.

So, $\varphi =\overrightarrow{E}\overrightarrow{A}cos{60}^0$

$\varphi =\frac{\sigma }{2{ϵ}_0}\times \pi ({10}^{-2}{)}^2\times \frac{1}{2}=17.5\frac{C}{N{m}^2}$