A cube is placed inside an electric field, E=150y2j^. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is:
(A)8.3×10−11C✓ Correct
(B)8.3×10−12C
(C)3.8×10−12C
(D)3.8×10−11C
Answer: (A)
Solution
The field E=150y2j^ points along y, so flux threads only the two faces perpendicular to y. The lower face lies in the plane y=0 where E=0 and contributes nothing. The upper face at y=0.5m has E=150(0.5)2=37.5N/C over area (0.5)2=0.25m2, giving flux Φ=37.5×0.25=9.375Nm2/C. By Gauss's law q=ε0Φ=8.85×10−12×9.375=8.3×10−11C.
JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option