The electric field in a region is given by $$\overrightarrow E = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right){N \over C}$$. The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y $-$ z plane) to that of the surface of area 0.3 m2 (parallel to x $-$ z plane) is a : b, where a = __________ [Here ${\widehat i}$, ${\widehat j}$ and ${\widehat k}$ are unit vectors along x, y and z-axes respectively.]
Answer (integer)
1
Solution
$\phi = \overrightarrow E \,.\,\overrightarrow A$<br><br>${\overrightarrow A _a} = 0.2\widehat i$<br><br>${\overrightarrow A _b} = 0.3\widehat j$<br><br>$${\phi _a} = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right).\,0.2\widehat i$$<br><br>$\Rightarrow$ ${\phi _a} = {3 \over 5}{E_0} \times 0.2$<br><br>$${\phi _b} = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right).\,0.3\widehat j$$<br><br>$\Rightarrow$ ${\phi _b} = {4 \over 5}{E_0} \times 0.3$<br><br>$${a \over b} = {{{\phi _a}} \over {{\phi _b}}} = {{{3 \over 5}{E_0} \times 0.2} \over {{4 \over 5}{E_0} \times 0.3}} = {6 \over {12}} = 0.5$$
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Field and Field Lines
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