# 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.

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

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.

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

= 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.

## 2 Answers

_{X}= x^(1/2) in this ‘x’ represents the distance from origin on x-axis. In figure you can see left face is at a distance of left face from origin is ‘a’ so,

_{X}(for left face) =

_{X}(for right face) =

Let us understand the problem in this way,

Electric Flux is defined as a number of electric field lines,

__passing per unit area__.

It is also defined as the product of electric field and surface area

__projected in a direction perpendicular to the electric field__.

As per given data electric field is in the x axis direction thats why we need to consider both the face perpendicular to x axis rest all are parallel to x axis. Hence no flux will develop.

Thanks

I hope above solution will clear your all doubts.

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All the best.