Question

The square surface shown in Fig. 23-30 measures 3.2mmon each side. It is immersed in a uniform electric field with magnitude E=1800 N/Cand with field lines at an angle of $\mathit{\theta }\mathbf{=}\mathbf{35}\mathbf{°}$with a normal to the surface, as shown. Take that normal to be directed “outward,” as though the surface were one face of a box. Calculate the electric flux through the surface.

Short answer

The electric flux through the surface is$-1.2\times {10}^{-2}\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.

Step-by-step solution

The given data

1. Side length of the square, a=32 mm
2. Electric field (E), E=1800 N/C
3. Field lines at an angle of $\theta =35°$with a normal

Understanding the concept of Gauss law-planar symmetry

Using the concept of the Gauss flux theorem, we can get the flux passing through the surface as per the given electric field. Now, the summation of all values will give the net flux through the volume.

Formula:

The electric flux through an enclosed surface,

$\mathrm{\varphi }=\overrightarrow{\mathrm{E}}.\overrightarrow{\mathrm{A}}=\mathrm{EAcos\theta }$ (1)

Calculation of the electric flux through the surface

The vector area A and the electric field E are shown in the diagram below:

The electric flux through the surface is given by

The angle between A and E is given as

180° – 35° = 145°

Therefore, the electric flux through the area is given using equation (i) as:

$$\begin{gathered} \varphi =\left(1800\mathrm{N}/\mathrm{C}\right){\left(3.2\times {10}^{-3}\mathrm{m}\right)}^2\cos \left(145°\right) \\ =-1.2\times {10}^{-2}\mathrm{N}.{\mathrm{m}}^2/\mathrm{C} \end{gathered}$$

Hence, the value of the electric flux is$-1.2\times {10}^{-2}\mathrm{N}.{\mathrm{m}}^2/\mathrm{C}$.