Question

Given the electric field in the region `vecE=2xhati`, find the net electric flux through the cube and the charge enclosed by it.

Answer 1

Since the electric field has only x component, for faces normal to x direction, the angle between E and ∆S is ±π/2. Therefore, the flux is separately zero for each face of the cube except the two shaded ones.

The magnitude of the electric field at the left face is E_L = 0      (As x = 0 at the left face)

The magnitude of the electric field at the right face is E_R = 2a    (As x = a at the right face)

The corresponding fluxes are

`phi_L=vecE.DeltavecS=0`

`phi_R=vecE_R.DeltavecS=E_RDeltaScostheta=E_RDeltaS " "(.:theta=0^@)`

⇒ϕR= ERa²

Net flux (ϕ) through the cube = ϕL+ϕR=0+ERa²=ERa²

ϕ=2a(a)²=2a³

We can use Gauss’s law to find the total charge q inside the cube.

`phi=q/(epsilon_0)`

q=ϕε₀=2a³ε₀