Hello,

This is example from Physics NCERT Book 12^th ! and i want to know how this Electric field is calulated for this both left and right faces !

Here is Question

Example 1.11 The electric field components in Fig. 1.27 are
Ex = áx1/2, Ey = Ez = 0, in which á = 800 N/C m1/2. Calculate (a) the
flux through the cube, and (b) the charge within the cube. Assume
that a = 0.1 m.

Sol. in NCERT :

Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between E and ΔS is
± π/2. Therefore, the flux φ = E.ΔS is separately zero for each face
of the cube except the two shaded ones. Now the magnitude of
the electric field at the left face is
EL = αx1/2 = αa1/2
(x = a at the left face).
The magnitude of electric field at the right face is
ER = α x1/2 = α (2a)1/2
(x = 2a at the right face).
The corresponding fluxes are
φL= EL
.ΔS = ΔS EL ⋅ nˆ L =EL ΔS cosθ = –EL ΔS, since θ = 180°
= –ELa2
φR= ER
.ΔS = ER ΔS cosθ = ER ΔS, since θ = 0°
= ERa2
Net flux through the cube

= φR + φL = ERa2 – ELa2 = a2 (ER – EL) = αa2 [(2a)1/2 – a1/2]
= αa5/2 ( 2 –1)
= 800 (0.1)5/2 ( 2 –1)
= 1.05 N m2 C–1
(b) We can use Gauss’s law to find the total charge q inside the cube.
We have φ = q/ε0 or q = φε0. Therefore,
q = 1.05 × 8.854 × 10–12 C = 9.27 × 10–12 C.

Let's break down the calculation of the electric field and the flux through the cube as described in your example from the NCERT Physics book. This involves understanding how the electric field behaves at different faces of the cube and applying Gauss's law to find the charge within the cube.

Understanding the Electric Field Components

In this scenario, the electric field is given as Ex = αx^(1/2), while Ey and Ez are zero. This means that the electric field only has a component in the x-direction, which simplifies our calculations significantly.

Calculating the Electric Field at Each Face

We need to evaluate the electric field at the two faces of the cube that are perpendicular to the x-axis: the left face and the right face.

• Left Face: At the left face, the position is x = a. Therefore, the electric field at this face is:

EL = αa^(1/2)

• Right Face: At the right face, the position is x = 2a. Thus, the electric field here is:

ER = α(2a)^(1/2)

Calculating the Flux Through the Cube

The electric flux φ through a surface is given by the equation:

φ = E · ΔS, where ΔS is the area vector of the surface.

Flux for Each Face

For the left face, the area vector points in the negative x-direction, making the angle θ between the electric field and the area vector equal to 180°. Thus, the flux through the left face is:

φL = EL · ΔS = -EL · ΔS

For the right face, the area vector points in the positive x-direction, so the angle θ is 0°. Therefore, the flux through the right face is:

φR = ER · ΔS = ER · ΔS

Net Flux Calculation

The total flux through the cube is the sum of the fluxes through the left and right faces:

Net Flux = φR + φL = ER · a² - EL · a²

Substituting the expressions for ER and EL gives us:

Net Flux = a² (ER - EL) = a² [α(2a)^(1/2) - αa^(1/2)]

This simplifies to:

Net Flux = αa² [(2)^(1/2) - 1] = αa^(5/2} (2 - 1)

Plugging in the values, we find:

Net Flux = 800 (0.1)^(5/2) = 1.05 N m²/C

Finding the Charge Inside the Cube

To find the charge q inside the cube, we can use Gauss's law, which states:

φ = q/ε₀ or rearranging gives us q = φε₀.

Substituting the flux we calculated and the permittivity of free space ε₀ ≈ 8.854 × 10⁻¹² C²/(N·m²):

q = 1.05 × 8.854 × 10⁻¹² C = 9.27 × 10⁻¹² C

In summary, we calculated the electric field at both faces of the cube, determined the flux through the cube, and applied Gauss's law to find the charge contained within it. This process illustrates the relationship between electric fields, flux, and charge in a clear and systematic way.