Challing Spherical Capacitor Problem

[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 a lot. 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?

 

[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?