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The conductivity ((sigma)) of a material can be calculated using the formula (sigma = J / E), where (J) is the current density and (E) is the electric field strength.

Given that:

– (J = 1) unit (assuming this is in A/m(^2) since current density is typically measured in Amperes per square meter)

– (E = 200) μV/m (since electric fields are typically measured in volts per meter, and (1) μV = (1 times 10^{-6}) V)

First, convert (E) to V/m for consistency: (200) μV = (200 times 10^{-6}) V/m

Now, we can calculate the conductivity:

[

sigma = frac{J}{E} = frac{1}{200 times 10^{-6}} = frac{1}{0.0002} = 5000 , text{S/m}

]

So, the conductivity of the material is (5000) Siemens per meter (S/m).

To determine if the given potential ( V = 25 sin theta ) satisfies Laplace’s equation in free space, we need to utilize the form of Laplace’s equation in spherical coordinates, as the potential is given in terms of (theta), which is a spherical coordinate.

Laplace’s equation in spherical coordinates (assuming azimuthal symmetry, which seems to be implied here as the potential is a function of (theta) alone) is:

[

nabla^2V = frac{1}{r^2}frac{partial}{partial r}left( r^2frac{partial V}{partial r} right) + frac{1}{r^2sintheta}frac{partial}{partial theta}left( sinthetafrac{partial V}{partial theta} right) + frac{1}{r^2sin^2theta}frac{partial^2 V}{partial phi^2} = 0

]

Given ( V = 25 sin theta ), this does not depend on ( r ) or ( phi ), so the first and third terms in the Laplace equation vanish. We’re only left with the second term:

[

frac{1}{r^2sintheta}frac{partial}{partial theta}left( sinthetafrac{

To compute Gauss’s law for a given electric displacement field ( mathbf{D} ) and to find the charge enclosed using a volume integral, we work step by step through the problem. Gauss’s law in differential form relates the divergence of the electric displacement field ( mathbf{D} ) to the free charge density ( rho_{free} ) present in the medium:

[

nabla cdot mathbf{D} = rho_{free}

]

Given:

[

mathbf{D} = frac{10rho^3}{4} hat{mathbf{i}}

]

This is given in cylindrical coordinates ((rho, phi, z)) but with a slight confusion in the notation since (hat{mathbf{i}}) is typically used for Cartesian coordinates. Assuming it’s meant to represent the radial component in cylindrical coordinates, it should correctly be (hat{rho}) instead of (hat{mathbf{i}}), so:

[

mathbf{D} = frac{10rho^3}{4} hat{rho}

]

To compute the charge enclosed within a cylindrical volume defined by ( rho = 4m ), between ( z = 0 ) and ( z = 5 ), we first compute the volume integral of the charge density ( rho_{free} ).

Since ( nabla

To solve this, we’ll use the concept of flux through a surface. The flux (Phi) of a vector field (textbf{D} = Ptextbf{i} + Qtextbf{j} + Rtextbf{k}) through a surface (S) is given by the surface integral of (textbf{D} cdot textbf{n} dS), where (textbf{n}) is the unit normal to the surface and (dS) is a differential element of the surface area.

Given (textbf{D} = 2xy textbf{i} + 3yz textbf{j} + 4xz textbf{k}) and considering the plane (x = 3) with (-1 < y < 2) and (0 < z < 4), we'll calculate the flux through this plane area.

For the plane (x = 3), the normal vector is parallel to the (textbf{i}) direction since the plane is perpendicular to the x-axis. Therefore, only the component of (textbf{D}) in the direction of (textbf{i}) contributes to the flux through this plane.

The relevant component of (textbf{D}) here is (P = 2xy), and since (x = 3), we have (P = 6y). The