Homework Help: Self capacitance of a sphere

1. Mar 11, 2013

Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
The self capacitance of an isolated sphere is $4\pi \epsilon_o R$ where R is the radius of sphere but I am not sure how to begin on this one.

Any help is appreciated. Thanks!

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2. Mar 11, 2013

voko

I do not think it is correct to talk about self-capacitance here. Self-capacitance involves a potential difference with a infinitely far away object, while in this case you are given two other objects at finite distances. I think you could treat the setup as two parallel capacitors, formed by spheres 1-2 and 2-3.

3. Mar 11, 2013

ehild

Add some charge Q onto the middle sphere, find the electric field and potential in the regions between the spheres. You know that the potential is zero on the inner and outer sphere.

ehild

4. Mar 11, 2013

Saitama

But how will that help in finding the self capacitance of the middle sphere?

Electric field outside the sphere is given by $kQ/r^2$ and inside it is zero but I am still clueless. :(

5. Mar 11, 2013

voko

You have ignored what I wrote earlier.

6. Mar 11, 2013

Saitama

Can you please shed some more light on this? I have checked your post more than 10 times but I still can't see what I have missed.

7. Mar 11, 2013

voko

You are explicitly given potentials between three objects, with finite distances from one another. Self-capacitance, however, is defined in terms of the potential difference with an infinite-radius sphere. You can't have it both ways.

8. Mar 11, 2013

Saitama

Do you mean that the given question is wrong then?

Continuing ehild's suggestion, if I give a charge to the middle sphere, will there be any charges induced on the other spheres?

9. Mar 11, 2013

ehild

Why should it be wrong?
Why not?

ehild

10. Mar 11, 2013

Saitama

Please see the attachment if I have mentioned the induced charges properly (I think they are wrong).

If they are right, should I calculate the potential at the surface of innermost and outermost spheres and set them equal to zero?

I did these types of problems in the past but now I have completely forgotten how to do them. :(

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11. Mar 11, 2013

SammyS

Staff Emeritus

That looks good to me, except for the charge on the outside of the outer sphere.

12. Mar 11, 2013

Saitama

Yes, I too think that its wrong that's why I wrote in brackets that they look wrong to me. Any hint about what should be those charges?

13. Mar 11, 2013

ehild

that is zero outside.

ehild

14. Mar 11, 2013

SammyS

Staff Emeritus
The potential of the sphere is zero, so the electric field exterior to the sphere must be zero.

15. Mar 11, 2013

Saitama

The charges mentioned on the inner surface of outermost sphere are right?

When I calculate the potential, do I need to add potential of both outer and inner surfaces of each sphere?

16. Mar 11, 2013

SammyS

Staff Emeritus
Yes. The charge on the inner surface are exactly opposite the charge on the outer surface of the middle shell.

No. (Assuming I understand your question correctly.)

The potential difference between the middle & outer shells should give you the charge, Q - q1 .

The potential difference between the inner & middle shells should give you the charge, q1 .

17. Mar 12, 2013

ehild

Apply Gauss's Law in the regions between the spheres. Integrate the electric field to get the potential difference between the middle sphere and the inner one, and also between the middle sphere and the outer one. They must be equal.

ehild

18. Mar 12, 2013

Saitama

The potential difference between the innermost and the middle sphere is:
$$kq_1 \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

The potential difference between the middle sphere and the outermost sphere is:
$$k(Q-q_1)\left(\frac{1}{R_3}-\frac{1}{R_2}\right)$$

Have I calculated the potential difference right?

19. Mar 12, 2013

ehild

Check the sign of the second one.

20. Mar 12, 2013

Saitama

I expected you to say that but I couldn't find the mistake in my working.
Potential at the surface of outer sphere is $k(Q-q_1)/R_3$ and potential at the surface of middle sphere is $k(Q-q_1)/R_2$, therefore the potential difference is $$k(Q-q_1)\left(\frac{1}{R_3}-\frac{1}{R_2}\right)$$.

21. Mar 12, 2013

ehild

The potential of both the inner and the outer sphere is zero (with respect to infinity). Assuming positive Q, the potential of the middle sphere is positive. What is the potential of the middle sphere?
You can not write the potential in this problem as kQ/r, as it is zero both at r=R1 and r=R3.
Do what I said about the electric field.

ehild

22. Mar 12, 2013

Saitama

The electric field between the middle sphere and the outermost sphere is $k(Q-q_1)/r^2$.
$$V(R_3)-V(R_2)=-\int_{R_2}^{R_3} \vec{E} \cdot \vec{dr}$$

$V(R_3)=0$. Hence
$$V(R_2)=k(Q-q_1)\left(\frac{1}{R_2}-\frac{1}{R_3}\right)$$

Is this right?

23. Mar 12, 2013

ehild

It is right. Now express V(R2) using the electric field between the innermost sphere and the middle one.

24. Mar 12, 2013

Saitama

$$V(R_2)-V(R_1)=-\int_{R_1}^{R_2} \frac{-kQ}{r^2} dr$$
Solving
$$V(R_2)=kq_1 \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

Equating both the expressions
$$(Q-q_1)\left(\frac{1}{R_2}-\frac{1}{R_3}\right)=q_1 \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
Solving,
$$q_1=\frac{Q\left(\frac{1}{R_2}-\frac{1}{R_3}\right)}{\left(\frac{1}{R_1}-\frac{1}{R_3}\right)}$$

Do I need to use $Q-q_1$ and the potential at the surface of the middle sphere in the equation $Q=C_{self}V$?

25. Mar 12, 2013

ehild

You have q1, and can plug that in to any equation for V(R2). The charge of the middle sphere is Q. q1 and Q-q1 are the surface charges.

ehild