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 areEx = áx1/2, Ey = Ez = 0, in which á = 800 N/C m1/2. Calculate (a) theflux through the cube, and (b) the charge within the cube. Assumethat a = 0.1 m.Sol. in NCERT :Since the electric field has only an x component, for facesperpendicular to x direction, the angle between E and ΔS is± π/2. Therefore, the flux φ = E.ΔS is separately zero for each faceof the cube except the two shaded ones. Now the magnitude ofthe electric field at the left face isEL = αx1/2 = αa1/2(x = a at the left face).The magnitude of electric field at the right face isER = α x1/2 = α (2a)1/2(x = 2a at the right face).The corresponding fluxes areφL= EL.ΔS = ΔS EL ⋅ nˆ L =EL ΔS cosθ = –EL ΔS, since θ = 180°= –ELa2φR= ER.ΔS = ER ΔS cosθ = ER ΔS, since θ = 0°= ERa2Net 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.

Mark Price , 8 Years ago

Grade 12th pass