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Given the Electric Field in the Region E=2xi,Find the Net Electric Flux Through the Cube and the Charge Enclosed by It. - Physics

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

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Solution

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 EL = 0      (As x = 0 at the left face)

The magnitude of the electric field at the right face is ER = 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

Concept: Electric Flux
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