Consider a cube with one corner at the origin and with sides of length 50 cm positioned along the x y z axes. There is an electric field E → = < 45, 210 y , 0 > N/C throughout the region that has a constant x component and a y component that increases linearly with y . (Assume the + x axis is to the right, the + y axis is upward, and the + z axis is out of the page. Also assume that the back left corner of the cube is at the origin.)

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!!!

he made 11 calls in one day