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प्रश्न
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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उत्तर
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
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संबंधित प्रश्न
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