Challing Spherical Capacitor Problem

[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[/color] 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?

 

[gneill]

The field outside the shell is the same as that for a point charge with a value equal to the sum of all the charges present - that of the inner sphere and the total for the shell. Add the inner sphere charge to the total charge on the shell. Call that Q. Stick the value into the field strength equation as before.

 

[mopar969]

So for r in the conductor E = 0
r1 < r < r2 E = 3.69 x10^4 Nm^2 /C all oiver r^2
r2 < r < r3 E = 5 (r^4 - r sub zero to the fourth) c/m^4 all over 4 epsilon zero r^2
r > r3 E = the previous electric field plus th electric field of r1 < r < r2.

Are all these correct and is this finally the end of the problem?

 

[gneill]

So for r in the conductor E = 0
r1 < r < r2 E = 3.69 x10^4 Nm^2 /C all oiver r^2
r2 < r < r3 E = 5 (r^4 - r sub zero to the fourth) c/m^4 all over 4 epsilon zero r^2
r > r3 E = the previous electric field plus th electric field of r1 < r < r2.

Are all these correct and is this finally the end of the problem?

The first and second are fine.

The third, for r2 < r < r3, should have r2 in place of "r sub zero", since that's where the shell begins. r₀ was a value I had used to indicate the beginning of the shell later in the discussion when I presented the integral. It is equivalent to r2, and r2 makes more sense in this context since it's used to delineate the regions.

For the fourth region (outside the whole ensemble), find the total charge, Q, for the whole ensemble! Then put that NUMBER into the expression for the field of a point charge, E = Q/(4 π ε₀ r²)

 

[mopar969]

So for the third region it is 5 (r^4 - r sub 2 to the fourth) c/m^4 all over 4 epsilon zero r^2

the fourth region is (3.69 x 10^4 Nm^2 /c all over r^2 + 5 pi (r^4 - r sub 2 to the fourth) c/m^4) all over 4 pi epsilon zero r^2

Does this reduce any further?

 

[gneill]

So for the third region it is 5 (r^4 - r sub 2 to the fourth) c/m^4 all over 4 epsilon zero r^2

the fourth region is (3.69 x 10^4 Nm^2 /c all over r^2 + 5 pi (r^4 - r sub 2 to the fourth) c/m^4) all over 4 pi epsilon zero r^2

Does this reduce any further?

*sigh* For the fourth region you haven't calculated the total charge! After the shell, the total charge no longer depends upon the radius; it is at its maximum, fixed, and final value.
This is where you can put the fixed radial limits of the shell into the integration and calculate the total charge of the shell, reducing it to a single number. Add it to the charge on the inner sphere and then plug that number into the expression for the field of a point charge to yield the expression for the field outside the shell.

 

[mopar969]

So for my r I should use 0.002 m and for r2 I should use 0.0012m and get a Q of 2.19 x10^ -10 couloumb add that to 7.4 x 10^-6 couloumbs and get 7.4 x10^-6 couloumbs. I then plug this into the electric field equation and got 6.65x10^4 nm^2/c all over r^2.

Is this correct? Also this is for r > r3 right? Also if I am correct are we finally done with this disaster of a problem?

 

[gneill]

Yes, yes, and I hope so

 

[mopar969]

One more thing what does the electric field in post 65 represent and for what region?

 

[gneill]

One more thing what does the electric field in post 65 represent and for what region?

That's the field inside the non-conducting shell.

 

[mopar969]

I thought that the electric field for that part was just 5 (r^4 - r sub two to the fourth) c/m^4 all over 4 epsilon zero r^2.

Since it is not what is this then?

 

[gneill]

The field in the shell is the sum of field sue to the inner sphere, and the field due to the portion of the shell interior to the radius. That is the two parts of that expression in post #65. What you've just quoted is only the part of the field contributed by the shell.

 

[mopar969]

Thanks so so much for the help gneill you are a life saver. By the way I am done now right? If you have time could you give a recap. Thanks

 

[gneill]

Thanks so so much for the help gneill you are a life saver. By the way I am done now right? If you have time could you give a recap. Thanks

You're welcome.

Yes, I believe you're done. Good luck.