Challing Spherical Capacitor Problem

[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.