Electrostatics

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297 of 311

Question 297 of 311

The electric field in a region is given by E⃗=(35E0i^+45E0j^)NC−1\vec{E}=\left(\dfrac{3}{5}{E}_{0}\hat{i}+\dfrac{4}{5}{E}_{0}\hat{j}\right)N{C}^{-1}. The ratio of flux of reported field through the rectangular surface of area 0.2m20.2{m}^{2} (parallel to y−zy-z plane) to that of the surface of area 0.3m20.3{m}^{2} (parallel to x−zx-z plane) is a:b=a:2,a:b=a:2, where a=?a=? [Here i^,j^\hat{i},\hat{j} and k^\hat{k} are unit vectors along x,yx,y and zz -axes respectively]

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Answer: 1

E⃗=(3E05i^+4E05j^)NC−1\vec{E}=\left(\dfrac{3{E}_{0}}{5}\hat{i}+\dfrac{4{E}_{0}}{5}\hat{j}\right)N{C}^{-1}
A1=0.2m2{A}_{1}=0.2{m}^{2} [parallel to y−zy-z plane]
=A⃗1=0.2m2i^={\vec{A}}_{1}=0.2{m}^{2}\hat{i}
A2=0.3m2{A}_{2}=0.3{m}^{2} [parallel to x−zx-z plane]
A⃗2=0.3m2j^{\vec{A}}_{2}=0.3{m}^{2}\hat{j}
Now, ϕa=[3E05i^+4E05j^]⋅[0.2i^]=3×0.25E0{\phi }_{a}=\left[\dfrac{3{E}_{0}}{5}\hat{i}+\dfrac{4{E}_{0}}{5}\hat{j}\right]\cdot \left[0.2\hat{i}\right]=\dfrac{3\times 0.2}{5}{E}_{0}
&{\phi }_{b}=\left[\dfrac{3{E}_{0}}{5}\hat{i}+\dfrac{4{E}_{0}}{5}\hat{j}\right]\cdot \left[0.3\hat{j}\right]=\dfrac{4\times 0.3}{5}{E}_{0}
Now, ϕaϕb=0.61.2=12=ab\dfrac{{\phi }_{a}}{{\phi }_{b}}=\dfrac{0.6}{1.2}=\dfrac{1}{2}=\dfrac{a}{b}
⇒a:b=1:2\Rightarrow a:b=1:2
⇒a=1\Rightarrow a=1

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