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Challing Spherical Capacitor Problem

by mopar969
Tags: capacitor, challing, spherical
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mopar969
#1
May1-11, 04:58 PM
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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ehild
#2
May2-11, 02:20 AM
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Quote Quote by mopar969 View Post
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
mopar969
#3
May2-11, 07:39 AM
P: 201
So for the hint Q / epsilon zero would be Q / epsilon zero times epsilon r. But what do I do next?

ehild
#4
May2-11, 08:13 AM
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Challing Spherical Capacitor Problem

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

ehild
mopar969
#5
May2-11, 08:17 AM
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?
mopar969
#6
May2-11, 10:42 AM
P: 201
Any suggestions to tackle this problem?
ehild
#7
May2-11, 10:50 AM
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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
#8
May2-11, 02:06 PM
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?
mopar969
#9
May2-11, 06:42 PM
P: 201
How do I use the er value?
mopar969
#10
May3-11, 06:14 AM
P: 201
I just need help putting the er value in?
mopar969
#11
May3-11, 09:00 AM
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.
mopar969
#12
May3-11, 10:25 AM
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?
mopar969
#13
May3-11, 10:35 AM
P: 201
Can I treat the spherical conductor of 7.4 micro coulombs as a point charge to find its electric field?
ehild
#14
May3-11, 11:21 AM
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Quote Quote by mopar969 View Post
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
gneill
#15
May3-11, 11:23 AM
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Quote Quote by mopar969 View Post
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

[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
#16
May3-11, 11:24 AM
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?
mopar969
#17
May3-11, 11:35 AM
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?
gneill
#18
May3-11, 11:45 AM
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Quote Quote by mopar969 View Post
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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