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The electric field in a region is given as:

\vec{E}=\frac{5}{4} E_0 \hat{i}+\frac{7}{3} E_0 \hat{j}+\frac{9}{2} E_0 \hat{k}

\\with\ E_0=3.0 \times 10^4 \mathrm{~N} / \mathrm{C}.\ \text {Calculate the flux of the field through rectangular surface of are}\ 2 \mathrm{~m}^2\ parallel\ to\ x-y\ plane\ in\ \mathrm{Nm}^2 / \mathrm{C}

Option: 1

2.7 \times 10^4


Option: 2

1.4 \times 10^5


Option: 3

2.7 \times 10^5


Option: 4

7.5 \times 10^4


Answers (1)

best_answer

\text { Area is parallel to } x-y \text { plane. }

\text {So, Area vector }(\vec{A})=0.2 \hat{k} \mathrm{~m}^2

\\From\ Gauss\ law, \\ \phi=\oint \vec{E} \cdot d A

=\left(\frac{5}{4} E_0 \hat{\imath}+\frac{7}{3} E_0 \hat{j}+\frac{9}{2} E_0 \hat{k}\right) \cdot(2 \hat{k})

=\frac{10}{4} E_0(i \cdot \hat{k})+\frac{14}{3} E_0(\hat{j} \cdot \hat{k})+\frac{18}{2} E_0(\hat{k} \cdot \hat{k})

=9 E_0 \quad\left[\begin{array}{l} \hat{i} \cdot \hat{k}=0 \\ \hat{j} \cdot \hat{k}=0 \\ k \cdot \hat{k}=1 \end{array}\right]

\begin{aligned} & =9 \times 3 \times 10^4 \mathrm{Nm}^2 / \mathrm{c} \\ & =27 \times 10^4 \mathrm{Nm}^2 / \mathrm{c} \\ & =2.7 \times 10^5 \mathrm{Nm}^2 / \mathrm{c} \end{aligned}

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