The electric field components are Ex=ax12,Ey=Ez=0, in which a=800N /Cm12. claculate the flux through cube and charge within cube assume a=0.1m. NCERT question on page 35 example 1.11 chapter 1 electric charges and fields.
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.
Answer:
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.