A spherical capacitor contains a solid spherical conductor of radius 0.5mm with

Question

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.)

I know that the electric field for a dielectric is e = Q all over 4 pi k epsilon zero r^2. I also know that the electric field for the innerest conductor can be found using E = 1 over 4 pi epsilon zero time Q over r^2. Using this equation I got 3.25 x 10 ^ -9. However I do not know how to solve the outer most shell given with the charge per volume value. And for my dielectric electric field formula how do I get the er value of 1.8 that was given in the problem into the equation?

Explanation / Answer

There are three regions: inside the conductor (i.e. r from 0 to 0.5 mm), inside the dielectric (r from 0.5 to 1.2 mm) and inside the outer shell (r from 1.2 to 2.0 mm)

Region 1: inside the conductor, E field is zero

Region 2: E = (1/k * epzero * 4pi) (Q / r^2) where k is the given dielectric constant of 1.8

Region 3: Now use Gauss' law...

(ep zero) integral of EdA = charge enclosed

choose some value of r, any random value between 1.2 and 2.0. Then the integral of EdA is just

E 4pi r^2

so you now have

(ep zero) 4pi r^2 * E = charge enclosed

the "charge enclosed" is the charge on the conductor in the middle, which I'll call Q, plus the charge enclosed in the shell within a sphere of radius "r". So...

charge enclosed = Q + integral of (charge density * dV)

We create dV from a spherical shell, which has area 4pi r^2 and thickness "dr". Then...

charge enclosed = Q + integral of ( 5r * 4pi r^2 dr)

and we integrate from r = 1.2 mm (which I will call "a") to the chosen "r"

So finally...

charge enclosed = Q + 20 pi * integral of (r^3 dr) =

= Q + 20pi ( 1/4 r^4) eval from a to "r" =

= Q + 5 pi (r^4 - a^4)

Put this back in to the right side of Gauss' Law above and...

(ep zero) 4pi r^2 E = Q + 5pi (r^4 - a^4)

Simplify and solve for E...

E = [ Q + 5pi (r^4 - a^4) ] / (ep zero) 4pi r^2

You can try to simplify this a little bit, but it's probably not worth the effort.

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