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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.

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 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.
 

Grade:12th pass

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