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Surface Integral

The electric ...

Question

The electric field in a region is given by $(\frac{3}{5} E_{\circ} \hat{i} + \frac{4}{5} E_{\circ} \hat{j})$ . The ratio of the flux of the reported field through the rectangular surface of area $0.2 m^{2}$ (parallel to y-z plane) to that of the surface of area $0.3 m^{2}$ (parallel to x-z plane) is $a : 2$ , where a = ________

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Solution

Step 1: Given data
Electric field $(\vec{E}) =$ $(\frac{3}{5} E_{\circ} \hat{i} + \frac{4}{5} E_{\circ} \hat{j})$
Area $(A_{a}) = 0.2 m^{2}$
Area $(A_{b}) = 0 \cdot 3 m^{2}$
Step 2: Formula used
The flux can be computed using the following formula:
$ϕ = \vec{E} \cdot \vec{A}$
Where $\vec{E} =$ electric field and $\vec{A} =$ Area
Step 3: Compute the flux $ϕ_{a}$
As the area is parallel to $Y - Z$ plane so normal unit vector to the surface is $\hat{n} = \hat{i}$ . So,
$ϕ_{a} = (\frac{3}{5} E_{\circ} \hat{i} + \frac{4}{5} E_{\circ} \hat{j}) \cdot (0 \cdot 2 \hat{i}) ϕ_{a} = \frac{3 \times 0 \cdot 2}{5} E_{\circ} ϕ_{a} = \frac{0 \cdot 6}{5} E_{\circ}$
Step 4: Compute the flux $ϕ_{b}$
As the area is parallel to $x - Z$ plane so normal unit vector to the surface is $\hat{n} = \hat{j}$ . So,
$ϕ_{b} = (\frac{3}{5} E_{\circ} \hat{i} + \frac{4}{5} E_{\circ} \hat{j}) \cdot (0 \cdot 3 \hat{j}) ϕ_{b} = \frac{4 \times 0 \cdot 3}{5} E_{\circ} ϕ_{b} = \frac{1 \cdot 2}{5} E_{\circ}$
Step 5: Compute the value of $a$
So, according to the problem the ratio of flux is,
$\frac{ϕ_{a}}{ϕ_{b}} = \frac{a}{b} \frac{0 \cdot 6}{5} \times \frac{5}{1 \cdot 2} = \frac{a}{b} \frac{1}{2} = \frac{a}{b}$
Hence, the value of $a$ will be $1$ .

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The electric field in a region is given by 35E∘i^+45E∘j^. The ratio of the flux of the reported field through the rectangular surface of area 0.2m2 (parallel to y-z plane) to that of the surface of area 0.3m2(parallel to x-z plane) is a:2, where a = ________