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JEE Main 2014 (19 Apr, Shift 1) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2014 (19 Apr, Shift 1)

Q113JEE Main 2014 · 19 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

The electric field in a region of space is given by, $\vec{E} = E_0\hat{i}$ + 2 $E_0\hat{j}$ where E0 = 100 N/C. The flux of this field through a circular surface of a radius 0.02 m parallel to the Y-Z plane is nearly:

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Answer: (a)

A surface parallel to the Y-Z plane has area vector along $\hat i$ , so only $E_x = E_0$ gives flux.
$\phi = E_0\pi r^2 = 100\times\pi\times(0.02)^2$
$\phi = 0.1257 \approx 0.125$ N m $^2$ /C, answer (a).

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