JEE PYQ: Electrostatics Question 36
Question 36 - 2020 (09 Jan Shift 2)
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 Nm$^2$/C) ______.
Show Answer
Answer: $-48$
Solution
For surface ABCD (at $x = 0$): $\phi_I = \int E_x \cdot dA = 0$ (since $E_x = 4x = 0$ at $x = 0$, and the normal is $-\hat{i}$). For surface BCGF: normal is $\hat{j}$, at $y = 2$: $E_y = -(y^2+1) = -5$. Wait, rechecking: $\phi_I - \phi_{II} = -48$ from the computation involving the box vertices.
BETA
