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Challing Spherical Capacitor Problem 
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#1
May111, 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 nonconducting 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
May211, 02:20 AM

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P: 10,634

As for the dielectric: read the hint. Replace ε_{0} with ε_{0} 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 


#3
May211, 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?



#4
May211, 08:13 AM

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P: 10,634

Challing Spherical Capacitor Problem
Apply Gauss' Law at any r to find the electric field strength. Do you know Gauss' Law?
ehild 


#5
May211, 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?



#6
May211, 10:42 AM

P: 201

Any suggestions to tackle this problem?



#7
May211, 10:50 AM

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P: 10,634

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=4πr^{2}dr. 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
May211, 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?



#9
May211, 06:42 PM

P: 201

How do I use the er value?



#10
May311, 06:14 AM

P: 201

I just need help putting the er value in?



#11
May311, 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.



#12
May311, 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?



#13
May311, 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?



#14
May311, 11:21 AM

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P: 10,634

0.5 mm<R<1.2 mm and use 1.8 ε_{0} instead of ε_{0}. In the range 1.2 mm<R<2 mm you have a nonconducting shell with a charge distribution. Hopefully, you can use ε_{0} 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: 4πR^{2}E=7.4 *10^{6} +4π∫(5r)r^{2}dr, 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 nonconducting 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
May311, 11:23 AM

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P: 11,685

[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] 


#16
May311, 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?



#17
May311, 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?



#18
May311, 11:45 AM

Mentor
P: 11,685

The charge density is specified to be "variable charge per unit volume p = 5r". Presumably that should yield units of C/m^{3}. Since the unit of distance we're using is mm, we can specify this charge density as 5r pC/mm^{4} (picocoulombs per mm^{4}, with r in mm). So your result should be in pico Coulombs (pC = 10^{12}C). 


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