Answer:
Charge in the cube = Q = (2.323 × 10⁻¹⁰) C
Explanation:
E → = < 45, 210 y , 0 > N/C
The length of the cube = 50 cm = 0.8 m
To find the charge in the cube, we need the net flux in the cube
In the z-direction, the electric field is 0, no, net flux in that direction,
In the x-direction, the electric field is constant, the same flux thay enters the face at x = 0 is the same that leaves at x= 0
50 cm, hence, no net flux in that direction too.
But the y direction, the electric field changes according to 210y from y = 0 to y = 0.50 m.
At y = 0, electric field = 210(0) = 0 N/C
At y = 0.50, electric field = 210(0.5) = 105 N/C.
Electric flux = Φ = E A = 105 (0.5 × 0.5) = 26.25 Vm or N/m²C
(the area of the xz plane that the field in this direction passes through is indeed 0.5²)
But according to Gauss' law, the net flux in a box is given by
Φ = Q/ε₀
where Q = charge in the box = ?
ε₀ = (8.85 × 10⁻¹²) C²N·m².
Q = Φε₀ = 26.25 × 8.85 × 10⁻¹²
Q = (2.323 × 10⁻¹⁰) C
Hope this Helps!!!
