Question

The electric field in a region is given by →E=35E0 →i+45 E0 →j with E0=2·0×103 N C-1. Find the flux of this field through a rectangular surface of area 0⋅2 m2 parallel to the y-z plane.

Answer 1

Answer:

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Answer 2

In the given plane surface, the flux of this field is 240Nm2/C.

Explanation:

According to the equation, the electric field:  $\frac{3}{5} E_{0} i+\frac{4}{5} E_{0} j$

E0=2.0 $\times$ 103 N/C

• Plane’s surface area = 0.2m2
• To the y-z plane, the plane is parallel. To the x-axis, this plane’s normal is parallel. So, the plane’s area vector is shown by 0.2m2î.

We know that,

$\phi=\vec{E} \cdot \Delta \vec{S}$

$\begin{aligned}&\Rightarrow \phi=\left(\frac{3}{5} E_{0} \hat{\imath}+\frac{4}{5} E_{0} \hat{\jmath}\right) \cdot(0.2 \hat{\imath})_{\mathrm{Nm}^{2} / \mathrm{C}}\\&\Rightarrow \phi=\frac{3}{5} \times 0.2 \times E_{0} \mathrm{Nm}^{2} \mathrm{C}\\&\left.\Rightarrow \phi=\frac{3}{5} \times 0.2 \times 2 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C} \text { (putting the value of } \mathrm{E}_{0}\right)\\&\phi=240 \mathrm{Nm}^{2 / \mathrm{C}}\end{aligned}$

Hence, the electric field’s flux via the provided surface of the plane is 240Nm2/C.