# Challing Spherical Capacitor Problem

by mopar969
Tags: capacitor, challing, spherical
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 P: 201 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?
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P: 10,671
 Quote by mopar969 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.
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
 P: 201 So for the hint Q / epsilon zero would be Q / epsilon zero times epsilon r. But what do I do next?
 HW Helper Thanks P: 10,671 Challing Spherical Capacitor Problem Apply Gauss' Law at any r to find the electric field strength. Do you know Gauss' Law? ehild
 P: 201 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?
 P: 201 Any suggestions to tackle this problem?
 HW Helper Thanks P: 10,671 To get the electric field inside the outer dielectric, calculate the the charge enclosed by a sphere of radius 1.2 mm
 P: 201 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?
 P: 201 How do I use the er value?
 P: 201 I just need help putting the er value in?
 P: 201 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.
 P: 201 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?
 P: 201 Can I treat the spherical conductor of 7.4 micro coulombs as a point charge to find its electric field?
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P: 10,671
 Quote by mopar969 Can I treat the spherical conductor of 7.4 micro coulombs as a point charge to find its electric field?
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
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P: 11,689
 Quote by mopar969 Can I treat the spherical conductor of 7.4 micro coulombs as a point charge to find its electric field?
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

$$E = \frac{Q}{4 \pi \epsilon_r\epsilon_0 r^2}$$
 P: 201 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?
 P: 201 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?
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P: 11,689
 Quote by 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?
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).

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