Challing Spherical Capacitor Problem

[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
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?

 

[gneill]

No you don't! The ONLY variable is the radius! When you solved the integral, the 'x' was a placeholder for the upper limit of integration -- it's the final radius value, whatever that might be, for the given radial position.

 

[mopar969]

Okay but where I am lost is that the question asks us to find the electric field everywhere so I do not know what the x value is?

 

[gneill]

The x value is everywhere within the shell! It's r, whatever r is, for any given location in the shell.

 

[mopar969]

I am lost with how you got the units for (6.65 x 104Nm2/C)/r2.?

 

[gneill]

What are the units for an electric field?

 

[mopar969]

I wanted to check something again with you:
When I integrated to calculate the charge per volume I got 5 pi r^4. Then I plugged it into
the charge equation in post # 47 and got 7.4 x 10 ^-6 C + 5 pi r^4. Is this correct? Thanks again for the help wit9h this problem.

 

[mopar969]

The units for electric field are N/C.

 

[gneill]

The units for electric field are N/C.

Right. (or V/m, which is the same thing).

So if you want your expression c/r², where c is a numerical constant and r is a distance in meters, to yield units for electric field, what units should you assign to the constant c?

 

[gneill]

I wanted to check something again with you:
When I integrated to calculate the charge per volume I got 5 pi r^4. Then I plugged it into
the charge equation in post # 47 and got 7.4 x 10 ^-6 C + 5 pi r^4. Is this correct? Thanks again for the help wit9h this problem.

That gives you the TOTAL CHARGE for the whole ensemble; the charge of the inner sphere plus the entire charge on the outer shell. It does not give you the charge enclosed within the Gaussian surface for partway through the shell. For that you need to perform the integration leaving the upper integration limit as a variable! That's the 'x' I've been trying to explain about. Using x for the variable name is only a matter of convenience to distinguish it from the variable r in the integration itself; it's a placeholder. Once the integration is done, you can replace x with the r variable.

Integrating out to radius x within the shell (the shell begins at radius r₀, the charge on the inner sphere is q₀):

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

$$Q(r) = 5 \pi (r^4 - r_0^4) C/m^4$$

 

[mopar969]

The Q = ... equation is that for r2> r > r3? Also, how do I plug this q into the electric field equation you gave a while back?

 

[gneill]

The Q = ... equation is that for r > r2 because the outer part is a shell. Also, how do I plug this q into the electric field equation you gave a while back?

It's for whatever range of radii comprise the shell. In the recent post I used r₀ as the start of the shell. Perhaps earlier I suggested r2. It's been so long that I'm losing track...

Q is a charge. Substitute the whole expression for Q in the E(r) = Q/(4πε₀r²) formula for the field.

 

[mopar969]

I got r2 from your picture which was a while ago. I got for the electric field 5 pi (r^4 - r sub zer to the fourth) all over 4 pi eo r^2. Does this reduce any further and how?

 

[gneill]

Don't forget the field due to the charge on the inner sphere.

If or why you would reduce the expression further depends upon your own esthetics and judgment. I would at least cancel the π's in the numerator and denominator.

 

[mopar969]

Do I just add the 7.4 x 10 ^-6 couloumbs to the end of the equation and what do you mean by n's?

 

[gneill]

Do I just add the 7.4 x 10 ^-6 couloumbs to the end of the equation and what do you mean by n's?

You add the FIELD due to that charge to the FIELD due to the shell's charge.

So E = q1/(4πε₀r²) + (5/4)*(r⁴ - r2⁴)/(ε₀*r²)(C/m⁴)

You can try to simplify further, but somehow I don't think it will get any better looking!

And the "n's" are π's, that is, they are the pi variable. The font used by PF does not render pi's very well!

 

[mopar969]

So the E(total) is:
5 (r to the fourth - r sub zero to the fourth)(c/m^4) all over 4 epsilon zero r^2 plus
3.69 x10^4 c/m^2 all over r^2.

Is this correct?

 

[gneill]

3.69 x10^4 Nm²/C/r² for that last bit. Always check that your units match what you're trying to find!

Otherwise, it looks like you're there for this portion of the problem.

 

[mopar969]

So what else is left for this problem?

 

[gneill]

If you've now got expressions or values for all the separate regions, you're done.

 

[mopar969]

But what about the region past the 2mm mark is it the last electric field equation (the E total) because it is a non conducting shell?