Question

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

Comments

Cos 180 = -1 then how is flux of left face 0??

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The magnitude of the electric field at the left face is E_L = 0, because x = 0 at the left face.

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

\(\phi _L = \vec E. \Delta \vec S = 0\)

\(\phi _R = \vec E_R.\Delta \vec S\)

\(= E_R \Delta S \cos \theta\)

\(= E_R \Delta S\)    \((\because \theta = 0°)\)

⇒ 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.

ϕ \(= \frac q{\varepsilon_0}\)

q = ϕε₀ = 2a³ε₀

Answer 2

Since, the electric field is parallel to the faces parallel to xy and xz planes, the electric flux through them is zero.

Electric flux through the left face, 

\(\phi_L = (E_L)(a^2)cos 180^o = 0\)

Electric flux through the right face, 
\(\phi_R = (E_R)(a^2)cos 0^o = (2a)(a^2)\times 1 = 2a^3\)

Net flux is given by, 

\(\phi = \frac{q_{enclosed}}{\in_0} = 2a^3\)

\(q_{enclosed} = 2a^3\varepsilon_o\)