Question

Given the net electric field in the region e=2xi find the net electric flux through the cube and the charge enclosed

Answer 1

The electric field in the horizontal direction, normally faces the cube which is makes an angle θ=90°, so the flux generated in such case is zero. The magnitudes of electric fields on right and left face of the cube is 3*a and 0.

So  EL = electric field magnitude on left face of cube; and ER = electric field magnitude on right face of cube. So the fluxes generated in the cube are;

ФL = EL* ΔS = 0  → be the equation A

ФR = EL *ΔS = ER*ΔS*cosθ

Since the angle θ= 0°, so cos 0°= 1 and putting that in above equation;

ФR = ER*ΔS*(1) = ER*ΔS → be the equation B

Net flux passing through the cube;

Фnet = ФL + ФR

So equation A & B implies that;

Фnet = 0 + ER*ΔS = ER*ΔS

where ΔS = a²

Фnet = ER*a²

⇒ q = a²*2*a = 2*a³

Now we can apply the Gauss' Law inside the cube structure which mathematically represents that;

Ф = q/ε₀

⇒ Ф = 2*a³/ε₀

Answer 2

Answer:

2a^3/epsilon not is your anwer

Explanation:

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