Components of electric field are given by Ex = alpha x, Ey = 0, Ez = 0. alpha is dimensionally constant. Calculate flux through each face of 'q' of side 'a' and effective charge inside 'q'.
Answers
Answer:
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Explanation:
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
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Given: Components of the electric field are given by Ex = alpha x, Ey = 0, Ez = 0. alpha is dimensionally constant.
To Find: Calculate flux through each face of 'q' of side 'a' and effective charge inside 'q'.
Solution:
The given electric field components are Ex = αx, Ey = 0, and Ez = 0, where α is a dimensionally constant.
To calculate the flux through each face of a cube q of side a, we can use Gauss's law. Gauss's law states that the flux through any closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (e0).
The electric flux through each face of the cube can be calculated as follows:
Flux through each face = Electric field component perpendicular to the face x Area of the face
For the cube q, the area of each face is a², and the normal vectors to each face are along the x, y, and z directions.
Therefore, the flux through the x-faces of the cube is given by:
x = Ex * A = αxa²
Similarly, the flux through the y-faces and z-faces of the cube is:
y = Ey * A = 0
z = Ez * A = 0
Since there is no electric field component in the y and z directions, the flux through the y-faces and z-faces is zero.
Now, to calculate the effective charge inside the cube, we can use Gauss's law again. The net flux through any closed surface is proportional to the charge enclosed by that surface.
The cube q is a closed surface, and the net flux through it is the sum of the fluxes through each face:
q = x + y + z = αxa²
According to Gauss's law, the net flux through the cube q is equal to the total charge enclosed by it divided by e0.
Therefore, the effective charge inside the cube q is:
q = q * e0 = αxa² * e0
So, the flux through each face of the cube q is x = αxa², and the effective charge inside the cube is q = αxa²e0.
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