To find the charge enclosed by a sphere of charge density ρ and radius a, you can use the formula for the charge, which is given by:

[ Q = rho cdot V ]

where ( V ) is the volume of the sphere. The volume ( V ) of a sphere is calculated using the formula:

[ V = frac{4}{3} pi r^3 ]

Substituting the radius ( a ) into the volume formula gives:

[ V = frac{4}{3} pi a^3 ]

Therefore, the total charge ( Q ) enclosed within the sphere is:

[ Q = rho cdot frac{4}{3} pi a^3 ]

So, the charge enclosed by the sphere is:

[ Q = frac{4}{3} pi rho a^3 ]

To apply the Divergence Theorem, we first need to compute the divergence of the vector field ( mathbf{D} = 2xy mathbf{i} + x^2 mathbf{j} ).

The divergence of a vector field ( mathbf{D} = P mathbf{i} + Q mathbf{j} + R mathbf{k} ) is given by:

[

nabla cdot mathbf{D} = frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z}

]

For our vector field ( mathbf{D} ):

– ( P = 2xy )

– ( Q = x^2 )

– ( R = 0 )

Now we compute the partial derivatives:

1. ( frac{partial P}{partial x} = frac{partial (2xy)}{partial x} = 2y )
2. ( frac{partial Q}{partial y} = frac{partial (x^2)}{partial y} = 0 )
3. ( frac{partial R}{partial z} = frac{partial (0)}{partial z} = 0 )

Thus, the divergence is:

[

nabla cdot mathbf{D} = 2y + 0 + 0 = 2y

To compute the divergence theorem for the vector field ( mathbf{D} = frac{5r^2}{4} mathbf{i} ) in spherical coordinates between ( r = 1 ) and ( r = 2 ), we follow these steps:

1. Convert the Vector Field to Spherical Coordinates:

In spherical coordinates, the relationship to Cartesian coordinates is given by:

– ( x = r sin theta cos phi )

– ( y = r sin theta sin phi )

– ( z = r cos theta )

The vector ( mathbf{i} ) represents the unit vector in the x-direction, so we express ( mathbf{D} ) as:

[

mathbf{D} = frac{5r^2}{4} mathbf{i} = frac{5r^2}{4} ( sin theta cos phi , sin theta sin phi , cos theta )

]

2. Compute the Divergence:

The divergence in spherical coordinates for a vector field ( mathbf{D} = (D_r, D_theta, D_phi) ) is given by:

[

nabla cdot mathbf{D} = frac{1}{r^2} frac{partial}{partial

To compute Gauss’s Law using the displacement field D, we first write down Gauss’s Law for electric displacement fields:

[

Phi_D = oint_S mathbf{D} cdot dmathbf{A} = Q_{free}

]

Where Q_free is the free charge enclosed by the surface S. Here, the displacement field is given as:

[

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

]

In cylindrical coordinates, D can be expressed as a function of ρ, and given constants indicate that D has only an i component (along the x-axis).

Given:

– ρ = 4 m

– z = 0 (which we could consider as the bottom face of a cylindrical volume)

– z = 5 (top face of the volume)

1. Calculate D at ρ = 4 m:

[

mathbf{D} = frac{10(4)^3}{4} mathbf{i} = frac{10 times 64}{4} mathbf{i} = 160 mathbf{i}

]

2. Define the Gaussian surface: A cylinder with radius 4 m and height 5 m.

3. Calculate the area vector: The enclosed area of the Gaussian surface includes the curved surface and two flat circular areas.

– Curved surface area