JEE Main 24 Feb 2021 Shift 2 Solved Paper

Given, charge, q=12µC=12×10−6C
Height of charge from surface, h=6cm=6×10−2m
and side of square, a=12cm=12×10−2m
From figure, it is clear that the given square is one of the face of a cube of side 12cm and +12µC charge is placed at its centre. Then, by Gauss's theorem,Flux through any face, φ=‌

q

6ε0

⇒‌‌φ‌‌=‌

12×10−6

6×8.854×10−12

=0.226×106N−m2∕C
‌‌=226×103N−m2∕C