Challing Spherical Capacitor Problem

mopar969

The 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.)





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?

ehild

You mean the electric field in the solid metal sphere? Is not it zero?

As for the dielectric: read the hint. Replace ε₀ with ε₀ ε_r and apply Gauss' Law systematically. For the outer part, determine the amount of charge embedded in the volume inside a sphere of radius r.

ehild

mopar969

So for the hint Q / epsilon zero would be Q / epsilon zero times epsilon r. But what do I do next?

ehild

Apply Gauss' Law at any r to find the electric field strength. Do you know Gauss' Law?

ehild

mopar969

I know that it says the enclosed integral of E times da is q over epsilon zero but I do not know how to apply it to this problem with a dielectric?

mopar969

Any suggestions to tackle this problem?

ehild

To get the electric field inside the outer dielectric, calculate the the charge enclosed by a sphere of radius 1.2 mm<R<2 mm. For that, you have to calculate the volume integral ∫ρdV. ρ is the charge density and it is given, ρ=5r. dV is the volume element. It can be replaced by the volume of a thin shell concentric with the innermost sphere, dV=4πr²dr. The volume integral gives the charge inside the outer dielectric, and you have to add the charge on the inner sphere to get all charge enclosed by a sphere of radius R.

ehild.

mopar969

When I did the integral 4 pi times 5/4 r^4 from 1.2 to 2 I got 218.755 couloumbs. I am a little confused as to what to do next and where the er value given comes in?

mopar969

How do I use the er value?

mopar969

I just need help putting the er value in?

mopar969

Can anybody help this problem is due today and I have been trying to work on it but am having no luck. PLEASE PLEASE PLEASE help.

mopar969

So, do I just add the 218.755 coulombs and the 7.4 micro coulombs together to get the total charge of the sphere or is there more to it?

mopar969

Can I treat the spherical conductor of 7.4 micro coulombs as a point charge to find its electric field?

ehild

Yes, do it in the dielectric, at distance R from the centre
0.5 mm<R<1.2 mm and use 1.8 ε₀ instead of ε₀.

In the range 1.2 mm<R<2 mm you have a non-conducting shell with a charge distribution. Hopefully, you can use ε₀ here, when using Gauss' Law: At a distance R from the centre, the surface integral of E over a sphere of radius R is equal to the charge enclosed by the sphere:
4πR²E=7.4 *10^-6 +4π∫(5r)r²dr, the integral goes from r=0.0012 to r=R.

For r>R, you can consider the total charge (that of the inner sphere + the charge of the non-conducting shell) as a point charge at the centre, and apply Coulomb's law to give the electric field strength as function of R.

ehild

gneill

Yes, provided that you take the radius at which you wish to know the field to be greater than the sphere's radius. Then

[tex] E = \frac{Q}{4 \pi \epsilon_0 r^2} [/tex]

If there's a dielectric material involved, then


[tex] E = \frac{Q}{4 \pi \epsilon_r\epsilon_0 r^2} [/tex]

mopar969

Okay but why did I get 218 couloumbs when I evaluated the integral for the area from 1.2 to 2 is this correct or did I mess up?

mopar969

Also when I set it as a point charge what will my r be will it be 1.2 or will it be 1.2 -0.5 = 0.7mm?