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]

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 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]

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]

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}

 

[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?

 

[gneill]

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).

 

[gneill]

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?

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

 

[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?

 

[gneill]

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?

The field strength for just within the outer shell looks good. The charge for the whole outer shell looks okay, too, if the given charge density is correct.

You may be required to find the field strength for any radial position within the outer shell, too, in which case you'll want to find an expression for the total charge contained within a given radius while "in" the shell. That means leaving the upper integration limit as a variable when you perform the integration.

 

[mopar969]

Wiith the charge I found how do I calculate the electric field do I just divide it by epsilon zero and I am little confused as to what to do next?

 

[gneill]

Wiith the charge I found how do I calculate the electric field do I just divide it by epsilon zero and I am little confused as to what to do next?

To find the field at radial distance r for a spherically symmetric charge distribution, apply the formula given previously:

<br /> E = \frac{Q}{4 \pi \epsilon_0 r^2} <br />

Where Q is the total charge enclosed within radius r.

(Be sure to include the contribution of ε_r if the radius ends within a dielectric).

 

[mopar969]

I got AN ANSWER OF 492301 n/c and I used 0.002 m for my r. Is this correct and how do I find the electric field past the 0.002 m mark because the problem says everywhere?

 

[gneill]

Your answer doesn't look correct for the field strength just outside the outer shell. Perhaps you should show how you calculated it.

 

[mopar969]

I divided 2.19 times ten to the -10 by 4 pi 8.85 times ten to the -12 and 0.002 m squared.

 

[gneill]

I see. That only accounts for the charge of the out shell. The charge on the inner sphere should be included -- remember that Gauss' law pertains to the net amount of charge contained withing the Gaussian surface.

 

[mopar969]

When I added up the charges I got 7.4 times ten to the -6 because the other charge was so small and got an electric field of 1.66 times ten to the tenth Newtons per couloumb. Is this correct and how do I calculate the electric field outside of the shell ??

 

[gneill]

When I added up the charges I got 7.4 times ten to the -6 because the other charge was so small and got an electric field of 1.66 times ten to the tenth Newtons per couloumb. Is this correct?

It is what it is!

Of course, this situation depends upon the charge density of the outer shell where we have assumed that the given relationship, namely 5r, meant that r was in meters and yielded values in units of Coulombs per cubic meter. If the implied units were otherwise, the situation could be different.

 

[mopar969]

Maybe I am not seeing how the sphere lokks correctly (If it is not too much can you plaese draw it). Also, So then how do you calculate for the electric field outside of the shell? Or is the 1.66 times ten to the tenth couloumbs the electric field for r > R. Thanks for all the help much appreciated.

 

[gneill]

The electric field of a spherically symmetric charge distribution, when viewed from outside, is identical to that of a point charge with the same value as the sum of all charge in the distribution located at the center of the sphere.

 

[ehild]

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?

It is wrong.

ehild

 

[mopar969]

Okay so my teacher wants the answers as a function of r sorry about all the tussel gneil so here is what I know the electric field inside the conductor is zero
the elctric field between (0.5 and 1.2) with respect to r is then 36949.6/r^2 N/C
the elctric field between (1.2 and 2) is 66539.4/r^2

I just need to know what the charge is beyond the 2mm and are these electric fields correct?

Also the carnot problem I posted the professor told me to do (1 - (tc/ti))(1-(ti/th)). So I am factoring that now but what do I do after I factor it.

Thanks again everybody I am learning alot. Thanks.

 

[gneill]

You should consider how many significant figures are appropriate for your answers.

For the region between 0.5 and 1.2 mm the result looks okay. But I question your result for the region inside the outer spherical shell. Firstly, the field value at its inner edge should be identical to the field value calculated for the dielectric region at that radius; the field can't have two values at the same location. Secondly, since within the region of the outer shell the net charge increases with radius out to the air boundary, the expression for the field should be more complex than a simple inverse square formula; You didn't heed my suggestion in post #21.

Once the radius exceeds the outer boundary of the non-conducting spherical shell the charge interior to the radius remains constant -- no more charge is added with distance. So then the field is simply that of a spherical charge with respect to radius.

 

[mopar969]

Do you know where I went wrong in my calculations. Thanks for the help?

 

[gneill]

Do you know where I went wrong in my calculations. Thanks for the help?

You need to derive the expression for the total charge interior to the radius for the region inside the outer spherical shell. It increases from the inner edge to the outer edge as radius increases. After that, the total charge remains constant with increasing distance.

 

[mopar969]

So how do I fix the 1.2 to 2 mm answer and how do I derive the equation for past the 2 mm. Sorry about this I am just bad at this type of problem.

 

[gneill]

Past the 2mm radius it is trivial: the field behaves as a point charge with the value of the charge being the total net enclosed charge due to the inner sphere and the outer shell.

Within the shell you need to sum the charge due to the inner sphere and the charge associated with the portion of the outer shell out to a given radius. The expression for that charge is then inserted into the formula for the field due to a spherical charge. You stated earlier that you solved the required integral for the whole shell thickness. Instead of the whole thickness, do it for an indefinite terminal radius.

 

[mopar969]

So for the whole shell tickness I used the equation e = Q total all over 4 pi epsilon zero r ^2. So what I did was not put a value in for r^2 and that is how I got 6.65x10^4 all over r^2.

 

[mopar969]

Maybe I messed up with the steps in my previous post?

 

[gneill]

Since the charge embodied in the outer shell seems to be so small compared with that on the inner sphere, it would appear that the latter charge dominates the result. So in this case, the field within the shell itself and outside the shell will follow the expression that you have, namely E = (6.65 x 10⁴Nm²/C)/r².

 

[mopar969]

So then my answers of
0
3.69 x 10^4/r^2 N/C
E = (6.65 x 104Nm2/C)/r2. for in the shell
and E = (6.65 x 104Nm2/C)/r2. for out the shell

are these correct and when I did my volume integral how come I did not set it up with respect to r? and for the 6.65 x 10 ^ 4 answer how come I did not need to use the er value given but I had to for the 3.69 x 10 ^ 4 answer?

 

[mopar969]

So then my answers of
0
3.69 x 10^4/r^2 N/C
E = (6.65 x 104Nm2/C)/r2. for in the shell
and E = (6.65 x 104Nm2/C)/r2. for out the shell

are these correct and when I did my volume integral how come I did not set it up with respect to r? and for the 6.65 x 10 ^ 4 answer how come I did not need to use the er value given but I had to for the 3.69 x 10 ^ 4 answer?

 

[gneill]

So then my answers of
0
3.69 x 10^4/r^2 N/C <--- need to account for m² due to r² [/color]
E = (6.65 x 104Nm2/C)/r2. for in the shell
and E = (6.65 x 104Nm2/C)/r2. for out the shell

are these correct and when I did my volume integral how come I did not set it up with respect to r? and for the 6.65 x 10 ^ 4 answer how come I did not need to use the er value given but I had to for the 3.69 x 10 ^ 4 answer?

The results look okay to me, although I'm still a bit concerned that the charge density of the shell was so small as to not effect the numerical results for the field calculations -- usually a carefully prepared example problem would not have this sort of issue. But, I suppose you have to deal with what's presented.

I don't recall having seen your volume integral workings. Your reported results, perhaps, but not the details of your integration steps.

The relative permittivity of the dielectric ε_r modifies the field only when the field is within the dielectric. Perhaps you might look at it as modifying the local properties of space, and effecting how the field presents itself to the local observer there.

 

[mopar969]

So then my answers of
0
3.69 x 10^4/r^2 N/C
E = (6.65 x 104Nm2/C)/r2. for in the shell
and E = (6.65 x 104Nm2/C)/r2. for out the shell

are these correct and when I did my volume integral how come I did not set it up with respect to r? and for the 6.65 x 10 ^ 4 answer how come I did not need to use the er value given but I had to for the 3.69 x 10 ^ 4 answer?

 

[mopar969]

My volume integral is the integral of row times dv where dv is 4 pi r^2 dr and row is 5r then I get 4 pi times ((5/4) r^4 from 0.0012 m to 0.002 m and I ggot a value of 2.19 x10^-10 couloumbs. Is this correct and I am lost on how to account for the m^2? because the equation I used Q all over 4 pi 1.8 epsilon zero r^2 does not have an m in it?

 

[gneill]

r is in meters so that r² has units of m². r is the only variable in your expressions, so that all the rest is a constant which must have the appropriate units so that the entire expression yields the correct units for field strength. Your units were fine for the other lines!

If you really want to have an accurate expression for the field inside the shell, you need to integrate over the radius between the inner surface of the shell and some indefinite location within the shell. In other words,

Q = q_0 + 20 \pi (C/m^4) \int_{r_0}^x r^3 dr

where r₀ is the radius of the inner surface of the shell (1.2 mm), and x is some radius that terminates somewhere within the shell. q₀ is the charge due to the inner sphere.

The units C/m⁴ are associated with the density expression's constant. Thus

ρ(r) = (5 C/m⁴)*r

to yield charge per unit volume.

 

[mopar969]

How do I use that integral t find the electric field then because it will have a variable?

 

[gneill]

The variable is the radial position. You want to find an expression for the field with respect to radius, right? So now you've got an expression for the charge with respect to radius. Insert the charge into the expression for the field of a charged sphere.

 

[mopar969]

That where I am lost because know I have more than just the variable r?