JEE PYQ: Electrostatics Question 2
Question 2 - 2021 (17 Mar Shift 2)
The electric field in a region is given by $\vec{E} = \frac{2}{5}E_0\hat{i} + \frac{3}{5}E_0\hat{j}$ with $E_0 = 4.0 \times 10^3$ N/C. The flux of this field through a rectangular surface area $0.4$ m$^2$ parallel to the $Y - Z$ plane is ______ Nm$^2$C$^{-1}$.
Show Answer
Answer: 640
Solution
$\phi = E_x \cdot A = \frac{2}{5} \times 4 \times 10^3 \times 0.4 = 640$ Nm$^2$/C.
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