The electric field in a region is given overr | vedclass

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Question

$\overrightarrow{ E }=\left(\frac{3 E _{0}}{5} \hat{ i }+\frac{4 E _{0}}{5} \hat{ j }\right) \frac{ N }{ C }$

$A_{1}=0.2 m ^{2}$ [parallel to $y -

Vedclass · Accepted Answer

$1$

Vedclass · Answer

$\overrightarrow{ E }=\left(\frac{3 E _{0}}{5} \hat{ i }+\frac{4 E _{0}}{5} \hat{ j }ight) \frac{ N }{ C }$

$A_{1}=0.2 m ^{2}$ [parallel to $y - z$ plane $]$

$=\overrightarrow{ A }_{1}=0.2 m ^{2} \hat{ i }$

$A_{2}=0.3 m ^{2}$ [parallel to $x - z$ plane $]$

$\overrightarrow{ A }_{2}=0.3 m ^{2} \hat{ j }$

Now $\phi_{a}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E_{0}}{5} \hat{j}ight] \cdot[0.2 \hat{i}]=\frac{3 	imes 0.2}{5} E _{0}$

$\phi_{b}=\left[\frac{3 E_{0}}{5} \hat{i}+\frac{4 E_{0}}{5} \hat{j}ight] \cdot[0.3 \hat{j}]=\frac{4 	imes 0.3}{5} E _{0}$

$\operatorname{Now} \frac{\phi_{ a }}{\phi_{ b }}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{ a }{ b }$

$\Rightarrow a: b=1: 2$

$\Rightarrow a=1$

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