An electric field E = 4xi - (y² + 1)j N/C passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as φI and φII respectively. The difference between (φI - φII) is (in Nm2/C) _______. Numerical answer type. Work it out, then open the answer.

Answer: -48. ABCD is the top face z = 2 (normal k); E has no z -component, so φI = 0. BCGF is the face x = 3 (normal i, area 2×2 = 4 m ²): Ex = 4(3) = 12, so φII = 48. φI - φII = 0 - 48 = -48 N m ² /C.

An electric field E = 4xi - (y² + 1)j N/C passes through the box… | JEE Main 2020 (9 Jan, Shift 2)

Q95JEE Main 2020 · 9 JAN · SHIFT 2Medium

Electric Flux & Gauss's Law

An electric field $\vec{E} = 4x\hat{i} - (y^{2} + 1)\hat{j}$ N/C

passes through the box shown in figure. The

flux of the electric field through surfaces ABCD

and BCGF are marked as $\phi_{I}$ and $\phi_{II}$

respectively. The difference between $(\phi_{I} - \phi_{II})$ is (in Nm2/C) _______.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: -48

ABCD is the top face $z = 2$ (normal $\hat k$ ); $\vec E$ has no $z$ -component, so $\phi_I = 0$ .
BCGF is the face $x = 3$ (normal $\hat i$ , area $2\times2 = 4$ m $^2$ ): $E_x = 4(3) = 12$ , so $\phi_{II} = 48$ .
$\phi_I - \phi_{II} = 0 - 48 = -48$ N m $^2$ /C.

Report a mistake

Electric Flux & Gauss's Law question from JEE Main 2020 (9 Jan, Shift 2)

More questions on Electric Flux & Gauss's Law