The electric field component are given as e x is equal to to a x raise to the power half E Y is equal Ez is equal to zero and Alpha is equal to 800 Newton per coulomb metre square calculate total flux through the cube and cha
rge enclosed in the cube assume a= 0.1 m
Answers
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