To compute the power consumed by the material, we can use the formula for power ((P)) in electrical systems, which is given by (P = IV), where (I) is the current in amperes (A) and (V) is the potential difference across the material in volts (V).

However, to find the current ((I)) flowing through the material, we first need to determine the total current from the given current density ((J)). Current density is defined as the current flow per unit area, given by (J = I/A), where (J) is the current density, (I) is the total current, and (A) is the area through which the current flows.

Given:

– Current density ((J)) = 15 units (assuming units are (A/m^2) for the purpose of this calculation, as actual units were not specified),

– Area ((A)) = 100 units ((m^2), assuming meters squared for coherence with current density units),

– Potential difference ((V)) = 20V.

We find the total current flowing ((I)) first:

[J = frac{I}{A} implies I = J times A]

[I = 15 , A/m^2 times 100 , m^2 = 1500 , A]

Then, we substitute (I) into

To find the current in the conductor, we need to understand the relationship between the given quantities. The electric field (E) in a conductor where an electric potential (V) is applied across a distance (d) can be expressed as (E = V / d). However, the electric field is given as 2 units (assuming SI units, this would be volts per meter), and the distance is 100 cm (which is 1 meter). From this, we can directly find the potential difference (V) across the conductor since (V = E times d).

Given:

– Electric field (E) = 2 V/m

– Distance (d) = 100 cm = 1 m

First, let’s find the potential difference (V):

[ V = E times d = 2 , text{V/m} times 1 , text{m} = 2 , text{V} ]

Now, Ohm’s law states that (V = IR), where (V) is the voltage across the conductor, (I) is the current through the conductor, and (R) is the resistance of the conductor. Given the resistance (R) is 2 ohms, we rearrange Ohm’s law to solve for the current (I):

[ I = frac{V}{R} ]

Substituting the given values:

[ I = frac{2 ,

To find the inductance of a coil, we can use the formula for the inductance of a solenoid, which is given by:

[L = mu N^2 A / l]

where:

– (L) is the inductance,

– (mu) is the permeability of the core material,

– (N) is the number of turns,

– (A) is the cross-sectional area,

– (l) is the length of the coil.

Given in the question:

– (mu = 3.5) (Assuming the unit here is Henrys per meter (H/m) since it’s about permeability and no specific unit is given),

– (N = 100) turns,

– The length ((l)) of the coil is (2m),

– The area ((A)) is thrice the length. However, to calculate area from length doesn’t directly compute without additional context, as length and area are not directly convertible. Assuming it means the side length of a square cross-section is thrice some basic unit, or perhaps a misunderstanding in the question’s phrasing. If it meant the area is proportional to the length in some manner not clearly defined, clarification is needed. Assuming a more typical approach where we define an area based off given or assumed dimensions: if by “thrice the length”, it means for example, the dimension contributing to the area is 3 times some value

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):