he question is:

A spherical capacitor contains a solid spherical conductor of radius 0.5mm with a charge of 7.4 micro coulombs, surrounded by a dielectric material with er = 1.8 out to a radius of 1.2mm, then an outer spherical non-conducting shell, with variable charge per unit volume p = 5r, with outer radius 2.0 mm. Determine the electric field everywhere. (Remember that in a linear dielectric material you can work out the equations as though they are in a vacuum and then replace e0 by e0er.)

To determine the electric field everywhere, we can approach this problem in several steps:

Step 1: Find the electric field inside the conductor.
Since the conductor is a solid sphere and the charge is uniformly distributed, the electric field inside the conductor is zero. This is because the charges on the surface of the conductor redistribute themselves in such a way that cancels out any electric field inside.

Step 2: Find the electric field in the dielectric material.
Since the dielectric material has a relative permittivity (er) of 1.8, we can use the equation for electric field in a linear dielectric:

E = E0 / er

where E0 is the electric field in a vacuum. Since we are given the charge inside the conductor (7.4 micro coulombs), we can use Gauss's law to find E0.

Gauss's law states that the electric flux through any closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε0).

∮E * dA = Q / ε0

Since the electric field outside the conductor is zero, we can consider a Gaussian surface enclosing only the solid spherical conductor. The electric flux through this surface is simply the electric field magnitude (E0) times the surface area of the sphere (4πr^2). The total charge enclosed is given as 7.4 micro coulombs. And since the permittivity of free space (ε0) is a known constant (8.854 x 10^-12 F/m), we can rearrange Gauss's law to solve for E0:

E0 = (Q / ε0) / (4πr^2)

Substituting the given values, we can calculate E0.

Step 3: Determine the electric field in the dielectric material.
Using the equation E = E0 / er, we can calculate the electric field in the dielectric material. The electric field is radial and points inward due to the positive charge on the conductor. So, the electric field in the dielectric material is given by:

E = E0 / er = (Q / ε0) / (4πr^2 * er)

Substituting the given values, we can calculate E.

Step 4: Find the electric field outside the dielectric material.
Since the outer shell is non-conducting and has a variable charge per unit volume (p = 5r), we can use a similar approach as Step 2 to find the electric field outside the dielectric material.

Consider a Gaussian surface just outside the dielectric material but inside the outer shell. The electric flux through this surface is equal to the electric field magnitude (E) times the surface area of the sphere (4πr^2). The total charge enclosed is given by integrating the charge density (p) over the volume of the shell.

Using Gauss's law, we can set up the equation:

∮E * dA = ∫p dv / ε0

Since the electric field is radial and points outward, the left-hand side of the equation simplifies to E * 4πr^2. The charge density (p) is given as 5r, and the volume element (dv) is equal to 4πr^2 dr (since it is a spherical shell). Thus, we have:

E * 4πr^2 = ∫(5r) * (4πr^2) dr / ε0

Integrating the right-hand side, we have:

E * 4πr^2 = (20π / 5ε0) * ∫(r^3) dr

E * 4πr^2 = (20π / 5ε0) * (r^4 / 4)

Simplifying, we get:

E = (5 / rε0) * (r^4 / 16)

E = (5r^3 / 16ε0)

Substituting the given values, we can calculate the electric field outside the dielectric material.

By following these steps, you can determine the electric field everywhere in the system.