Gauss' Law

1. Jun 8, 2008

yahoo32

We have two thin concentric conducting spheres r1 and r2. THe spheres are connected with a power source which supplies emf of epsilon_0. The smaller sphere is connected to the positive pole of the power supply and the larger sphere is connected to the negative pole. Using Gauss Law find the electric field between r1< R <r2 and find the electric field R>r2.

For this the electric field R. r2 would be 0 correct? Because charge enclosed is 0?

And for the electric field between r1 and r2 it would be equal to E = (+Q)/(4pi*R^2*Epsilon_0)?

After disconnecting the battery we ground the outer shell. Now calculate the electric field R>r2 and R<r1.

For R<r1, the electric field is 0 because charge enclosed is 0, correct?

And for R>r2 the electric field is now equal to E = (+Q)/(4pi*R^2*Epsilon_0) because charge enclosed is now +Q +Q -Q = +Q.

Could you guys please verify if my reasoning is correct for these problems. Thank you so much for your help!

2. Jun 8, 2008

anyone?

3. Jun 8, 2008

EngageEngage

Since the smaller sphere acquires a positive potential, there will be an outward electric field between the spherical shells. Your task should be to find out what charge will accumulate on the spheres as a result of the potential difference. How can you relate charge to potential?

4. Jun 8, 2008

yahoo32

We are given Q. I just need to show what the electric field would be in terms of Q. Therefore in between the spheres, as you stated, the electric field would decrease as a function of 1/r^2, correct? Could you state which part of the problem you are referring to? Because once you ground it, the positive charge should leave from the sphere correct?

5. Jun 8, 2008

EngageEngage

You didnt state that you had Q at the beginning of the problem> I was referring to the very first part where you're supposed to find the E field between the spheres. The E field there is not zero if there is a potential applied between the spheres.
EDIT: ahh, didnt see that you had that part, but it looks good.
EDIT2:Thats what I would have done, unless I missed something.

Last edited: Jun 8, 2008
6. Jun 8, 2008

yahoo32

Sorry. So Q is given to me. Now would it be true that for the first part the electric field outside the concentric spheres is 0 and between them is equal to E = (+Q)/(4pi*R^2*Epsilon_0)?

7. Jun 8, 2008

EngageEngage

Yeah, that looks right since the superposition of charges in gauss' law is a linear operation, your net charge is zero outside the second shell. inbetween the only charge contributing to electric flux is contained on the inner sphere (which terminates on the outer sphere)

8. Jun 8, 2008

yahoo32

I see. So is my equation correct for the charge in between the two conducting concentric spheres? Also could you also verify my reasoning for when the outershell is grounded?

9. Jun 8, 2008

EngageEngage

Actually, wait I think we missed something. If there is a potential between the spheres, your eq is:

V = q/(4piepsilon)*(1/ra - 1/rb)
and your E field between the spheres becomes
E = V/((1/ra - 1/rb)*r^2)

it seems redunant that they give you charge and the potential between the spheres.

10. Jun 8, 2008

EngageEngage

When the outer shell is grounded, it has access to any charge flow, which will be determined by the charge on the smaller sphere. Once the power is disconnected (as long as the potential is not gradually lowered, if thats possible) the spheres retain their charge i believe. So, after grounding the larger shell, its charge is determined by the charge that is on the smaller sphere, which will 'pull' an equal amount of charge towards the inner surface of the shell. What happens after this?

11. Jun 8, 2008

yahoo32

It is asking to calculate it using Gauss' Law. And after you ground the outer shell, the total charge enclosed at r=20cm is +Q, correct? Wait that new formula you calculated for E, is that really true? That would throw everything off in this last part of the problem?

12. Jun 8, 2008

EngageEngage

The formula will reduce to what you have if q is actually Q on the inner shell -- im sort of confused why they give you both Q and the potential difference applied between the shells. It seemed to me first that this was a spherical capacitor problem because of this.
Edit: if it is not Q, though, it wont be hard to calculate using the formula for V above. But, your general formula will be right because the charge on a spherical conductor acts as if its concentrated at a point in the middle of a sphere.

Last edited: Jun 8, 2008
13. Jun 9, 2008

yahoo32

I am very confused now. There was a portion of this problem I solved before that required EMF to solve because it asked to solve for the charge and the energy stored in the electric field. Maybe the potential difference isn't required for this part?

14. Jun 9, 2008

EngageEngage

Ahh, ok, well forget what I've been saying then. If you've solved for the charge and have set it to Q, then your formula for between the sphere in the first part is correct.

15. Jun 9, 2008

yahoo32

Okay, how about the part of the problem where we ground it. Is my reasoning correct for those two electric fields as well?

16. Jun 9, 2008

EngageEngage

This one, however, I'm not sure if you have the right answer, since the outer conductor is grounded. (this would be true if the conductor was neutral, and ungrounded). Since the conductor is grounded,it can pick up net charge because it has unlimited charge to take up from the 'earth'. so, if the inner conductor is at +Q, the outer conductor will match this with -Q on its inner surface. In a neutral conductor, the outer surface would become +Q, to keep the net charge at 0. However, the conductor is now grounded which leads me to believe that the outer conductor will attain a net charge of -Q, changing your answer. IM NOT 100% on this.

17. Jun 9, 2008

yahoo32

I see, could anyone else provide their opinion? The way I see it, you lose the -Q charge meaning Q enclosed at a distance outside both spheres would be +Q -Q +Q which is equal to Q. Anyone?

18. Jun 9, 2008

rl.bhat

When you ground the outer sphere, there will be no charge on the outer surface of the outer sphere. But there will be negative charge on the inner surface of the outer sphere because of the positive charge on the outer surface of the inner sphere. And there is no charge on the inner surface of the inner sphere. So there is no electric field at R<r1 and R>r2.

19. Jun 9, 2008

yahoo32

Wait a sec, if it is true that once you ground it there is no electric field at R.r2, then that would mean it is also true that when it is NOT grounded there IS an electric field at R>r2, which is not true, according to what I think...

20. Jun 9, 2008

rl.bhat

When you connect a power supply to a neutral concentric spheres, the electrons were transferred from innere sphere to outer sphere. Since charges always reside on the outermost surface of the conductor, outer surfaces of inner and outer spheres have equal and opposite charges. Hence field at R>r2 is zero. When you connect the outer sphere to the ground, the electrons move to the ground. Now there is no charge on the outer surface of the outer sphere. At this instant the +ve charge on the inner sphere induces equal amount of electrons on the innere surface of the outer sphere. Equal amoungt of +ve charges are induced on the outer surface of the outer sphere which are neutralised by the supply of electrons from the ground. In this case also electric field is zero at R>r2.
If you remove the ground connection, electrons on the inner surface will move toward the outer surface. And that will be the same case as in the first one.

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