JEE PYQ: Electrostatics Question 7
Question 7 - 2021 (25 Feb Shift 1)
The electric field in a region is given by $\vec{E} = \left(\frac{3}{5}E_0\hat{i} + \frac{4}{5}E_0\hat{j}\right)\frac{\text{N}}{\text{C}}$. The ratio of flux of reported field through the rectangular surface of area $0.2$ m$^2$ (parallel to $y - z$ plane) to that of the surface of area $0.3$ m$^2$ (parallel to $x - z$ plane) is $a : b$, where $a$ = ______ (round off to nearest integer)
Show Answer
Answer: (1)
Solution
$\phi_a = \frac{3}{5}E_0 \times 0.2$ and $\phi_b = \frac{4}{5}E_0 \times 0.3$. Ratio $= \frac{3/5 \times 0.2}{4/5 \times 0.3} = \frac{0.12}{0.24} = \frac{1}{2} = 0.5$. So $a = 1$ (ratio is $1:2$, $a = 1$).
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