To find the electric flux density surrounding a material with a given electric field intensity in transformer oil, we first need to understand the relationship between electric flux density ((mathbf{D})), electric field intensity ((mathbf{E})), and the permittivity of the medium ((epsilon)). The relationship can be expressed as:

[mathbf{D} = epsilon mathbf{E}]

where (epsilon = epsilon_0 epsilon_r), (epsilon_0) is the permittivity of free space ((8.85 times 10^{-12} F/m)), and (epsilon_r) is the relative permittivity of the medium (for transformer oil, (epsilon_r = 2.2)).

Given:

– Electric field intensity, (mathbf{E} = 2xyz) (where (x), (y), and (z) are the coordinates in meters),

– Relative permittivity of transformer oil, (epsilon_r = 2.2),

– Point (P(1,2,3)).

First, calculate (epsilon):

[epsilon = epsilon_0 epsilon_r = (8.85 times 10^{-12} F/m) times 2.2]

[epsilon = 1.947 times 10^{-11} F/m]

Next, calculate the electric field

To find the electric flux density (D) at (R = 6m) for a configuration involving multiple charged cylindrical sheets, we employ the concept of Gauss’s law. Gauss’s law in the context of electric flux density states that the net electric flux through a closed surface is equal to the charge enclosed by that surface.

Given:

1. (sigma = 5 , text{C/m}^2) at (R = 2m)
2. (sigma = -2 , text{C/m}^2) at (R = 4m)
3. (sigma = -3 , text{C/m}^2) at (R = 5m)

And we are asked to find the flux density (D) at (R = 6m).

To start, we calculate the charge per unit length (lambda) on each cylindrical sheet using the surface charge density (sigma) and the relation (lambda = sigma cdot 2pi R) because the charge per unit length for a cylindrical shell of radius (R) is the product of the surface charge density and the circumference of the cylinder.

1. For (R = 2m): (lambda_1 = 5 cdot 2pi (2) = 20pi , text{C/m})
2. For (R = 4m):

To find the electric flux density (D) at (R = 3m) in the arrangement of charged cylindrical sheets described, we use Gauss’s law in its integral form which relates the electric flux through a closed surface to the charge enclosed by that surface. For a cylindrical symmetry, the electric field (E) (and thus the electric flux density (D), since (D = epsilon_0 E) in vacuum or air where (epsilon_0) is the permittivity of free space) at a distance (R) from the axis is due to the enclosed charge per unit length ((lambda)) divided by (2piepsilon_0 R).

Given, we have three cylindrical sheets:

– The first with a surface charge density (sigma = 5 , text{C/m}^2) at (R = 2m),

– The second with (sigma = -2 , text{C/m}^2) at (R = 4m),

– The third with (sigma = -3 , text{C/m}^2) at (R = 5m).

Given charges are surface charge densities on cylindrical sheets; thus, the enclosed charge by a Gaussian cylinder of radius (R = 3m) only includes the charge from the first cylindrical sheet ((R = 2m)).

To calculate the total enclosed charge per unit length (

To find the electric potential (V) at a point due to a charged ring, we can use the formula for the potential due to a ring of charge. Given a ring with linear charge density (lambda) (charge per unit length) and radius (R), and a point along the axis perpendicular to the plane of the ring and passing through its center at a distance (x) from the center of the ring, the electric potential (V) at that point is given by:

[ V = frac{kQ}{sqrt{R^2 + x^2}} ]

where (k) is Coulomb’s constant ((k approx 8.987 times 10^9 , text{Nm}^2/text{C}^2)), (Q) is the total charge on the ring, (R) is the radius of the ring, and (x) is the distance of the point from the center of the ring along the axis. The total charge (Q) can be found from the charge density (lambda) and the circumference of the ring ((Q = lambda cdot 2pi R)).

Given that the linear charge density (lambda = 2) units and the radius of the ring (R = 2) meters, and the point is at a distance of (x = 1) meter from the center of the ring along its axis, let’s calculate the potential

The net electric flux, (Phi), crossing a surface that encloses a charge is given by Gauss’s law. Gauss’s law states that the flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space ((epsilon_0)). Mathematically, Gauss’s law is expressed as:

[

Phi = frac{Q_{text{enc}}}{epsilon_0}

]

The charge density is given as a function of (phi), (rho_s = 10sinphi), where (rho_s) represents the surface charge density. The total charge enclosed, (Q_{text{enc}}), can be calculated by integrating the charge density over the entire disc. The surface area element for a disc in polar coordinates is (dA = r,dr,dphi), where (r) is the radius and (phi) is the polar angle. However, since the charge density is given as a function of (phi) without dependence on (r), and considering the circular symmetry and the nature of the sine function, we can analyze this considering the charge distribution’s nature over ([0, 2pi]) range of (phi).

For a disc with a radius of 5m, integrating the charge density over the surface should account for the entire area, but the charge density varies with the angular component