# Challing Spherical Capacitor Problem

1. May 1, 2011

### 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?

2. May 2, 2011

### 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 εr 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

3. May 2, 2011

### mopar969

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

4. May 2, 2011

### ehild

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

ehild

5. May 2, 2011

### 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?

6. May 2, 2011

### mopar969

Any suggestions to tackle this problem?

7. May 2, 2011

### 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.

8. May 2, 2011

### 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?

9. May 2, 2011

### mopar969

How do I use the er value?

10. May 3, 2011

### mopar969

I just need help putting the er value in?

11. May 3, 2011

### mopar969

12. May 3, 2011

### 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?

13. May 3, 2011

### mopar969

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

14. May 3, 2011

### 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

15. May 3, 2011

### Staff: Mentor

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

$$E = \frac{Q}{4 \pi \epsilon_0 r^2}$$

If there's a dielectric material involved, then

$$E = \frac{Q}{4 \pi \epsilon_r\epsilon_0 r^2}$$

16. May 3, 2011

### 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?

17. May 3, 2011

### 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?

18. May 3, 2011

### Staff: Mentor

I think that you may have an order-of-magnitude issue due to the units involved.

The charge density is specified to be "variable charge per unit volume p = 5r". Presumably that should yield units of C/m3. Since the unit of distance we're using is mm, we can specify this charge density as 5r pC/mm4 (pico-coulombs per mm4, with r in mm).

So your result should be in pico Coulombs (pC = 10-12C).

19. May 3, 2011

### Staff: Mentor

Radius r is always measured from the sphere's center.

20. May 3, 2011

### mopar969

Okay so the r is 0.0012meters and when I calculated the electric field I got 2.56 x10^10 and for the charge per volume part I changed my mm to m and got 2.19 x10 ^ -10 couloumbs are these numbers looking correct now or no because that seems small to me?