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  • The electric field in a region is given by <img alt="\overrightarrow{E} = \left ( \frac{3}{5}E_0\widehat{i}+\frac{4}{5} E_0 \widehat{j}\right )\frac{N}{C}" src="https://entrancecorner.codecogs.com/gif.latex?%5Coverrightarrow%7BE%7D%20%3D%20%5Cleft

The electric field in a region is given by \overrightarrow{E} = \left ( \frac{3}{5}E_0\widehat{i}+\frac{4}{5} E_0 \widehat{j}\right )\frac{N}{C}. the ratio of flux of reported field through the rectangular surface of area 0.2m2 (parallel by y-z plane) to that of the surface of area 0.3m2 (parallel by x-z plane) is a:b , where a =________ [Here \widehat{i},\widehat{j} \ and\ \ \widehat{k} are unit vectors along x,y and z respectively]
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 4

Answers (1)

best_answer

\begin{aligned} &\overrightarrow{\mathrm{E}}=\left(\frac{3 \mathrm{E}_{0}}{5} \hat{\mathrm{i}}+\frac{4 \mathrm{E}_{0}}{5} \hat{\mathrm{j}}\right) \frac{\mathrm{N}}{\mathrm{C}}\\ &\mathrm{A}_{1}=0.2 \mathrm{~m}^{2} \text { [parallel to } \mathrm{y}-\mathrm{z} \text { plane }]\\ &\Rightarrow \vec{A}_{1}=0.2 \mathrm{~m}^{2} \hat{\mathrm{i}}\\ &\mathrm{A}_{2}=0.3 \mathrm{~m}^{2} \text { [parallel to } \mathrm{x}-\mathrm{z} \text { plane }]\\ &\Rightarrow \overrightarrow{\mathrm{A}}_{2}=0.3 \mathrm{~m}^{2} \hat{\mathrm{j}} \end{aligned}

\begin{array}{l} \text { Now } \phi_{\mathrm{a}}=\left[\frac{3 \mathrm{E}_{0}}{5} \hat{\mathrm{i}}+\frac{4 \mathrm{E}_{0}}{5} \hat{\mathrm{j}}\right] \cdot[0.2 \hat{\mathrm{i}}]=\frac{3 \times 0.2}{5} \mathrm{E}_{0} \\ \\ \& \quad \phi_{\mathrm{b}}=\left[\frac{3 \mathrm{E}_{0}}{5} \hat{\mathrm{i}}+\frac{4 \mathrm{E}_{0}}{5} \hat{\mathrm{j}}\right] \cdot[0.3 \hat{\mathrm{j}}]=\frac{4 \times 0.3}{5} \mathrm{E}_{0} \end{array}

\begin{array}{l} \text { Now } \frac{\phi_{a}}{\phi_{b}}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{a}{b} \\ \\ \Rightarrow a: b=1: 2 \\ \Rightarrow a=1 \end{array}

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