Question bank/JEE Main/Physics

An electric field
$\overset{\rightarrow}{E} = 4x\overset{\hat{}}{i} - (y^{2} + 1)jN/C$
passes through the box shown in figure. The flux of the electric field
through surface ABCDand BCGF and marked as $\phi_{1}$ and $\phi_{2}$
respectively. The difference between
$(\phi_{2} - \phi_{1})is\ (in{Nm}^{2}/C)$
[IMAGE] fudyrk gSA;fn cD ABCD rFkk BCGF leryksa

A47B46C42D48

Show answer and solution▾

Flux via ABCD ls ¶yDl
$\phi_{1} = \int\overset{\rightarrow}{E} \cdot d\overset{\rightarrow}{A} = 0$
[IMAGE] Flux via BCEF ls ¶yDl
$= 16x,x = 3$
$\phi_{2} = 48\frac{N - m^{2}}{C};\phi_{2} - \phi_{1} = 48\frac{N - m^{2}}{C}$

Source: Grass-G / P-21 Physcis Paper 19 Dec Hindi Eng..docx