Follow up on the 2 spheres capacitor problem. Help needed before tomorrow

In summary, the conversation discussed the problem of finding the capacitance of two isolated conducting spheres of equal radius with opposite charges when separated by a large distance. The formula for capacitance was derived and a potential difference equation was used to find the capacitance. It was found that the calculated capacitance was 0, which seemed incorrect. A Taylor expansion was suggested as a possible solution, which resulted in a negative capacitance. It was then discussed that the Taylor expansion was a good idea and the correct answer was obtained by taking into account the ratio of R to L. The final formula for capacitance was found to be C = R/2k + R^2/2kL.
  • #1
Lisa...
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0
Follow up on the 2 spheres capacitor problem. Help needed before tomorrow!

It was probably a bit hard to see what the old topic was about, so I started a new one. I'm still not quite sure on this problem, so I'd REALLY appreciate it if somebody could clear things up a little bit more. The problem is the following one:

Two isolated conducting spheres of equal radius have charges +Q (#1)and -Q (#2) respectively. If they are separated by a large distance compared to their radius, what is the capacitance of this unusual capacitor?

I've drawn this picture:

http://img147.imageshack.us/img147/7725/naamloos6ha.gif [Broken]

The origin is in the center of the left sphere.


This is what I've found till now:

[tex]C= \frac{Q}{V}[/tex]
In this case Q= Q = the absolute charge on every sphere.
V= the potential difference between #1 and #2, given by [tex]V= - \int_{#1}^{#2} E dl = = \int_{#2}^{#1} E dl[/tex]
with dl= dxi +dyj+dzk

The electric field due both spheres is determined by Coulombs law ([tex] E= \frac{kQ}{r^2}[/tex]).
Pick a point at a distance r from the origin somewhere on the drawn x-axis. In that point the E field of #1 and #2 is felt in the positive x direction (#1 is positive, so the field lines take off from #1 and #2 is negative, so the field lines approach #2). Therefore the magnitudes of the electric fields of #1 and #2 add and the resultant E is in the positive x direction.

The distance from the point P to the center of #1 (in the origin)= r.
Therefore [tex] E_1= \frac{kQ}{r^2}[/tex]
The distance from the point P to the center of #2 = [L+2R]-r.
Therefore [tex] E_2= \frac{kQ}{([L+2R]-r)^2}[/tex]
And [tex] E total= \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2}[/tex]

Now substitute this in the integral [tex]V= \int_{#2}^{#1} E dl [/tex] with values of:

[tex] E= E total= \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2}[/tex]
dl= dr, with dr the projection of dl in radial direction and
#1= R
#2= L+R

(that is the domain of the space between the two spheres where the point P can be chosen in)

This gives:

[tex]V= \int_{L+R}^{R} \frac{kQ}{r^2} + \frac{kQ}{([L+2R]-r)^2} dr = \left[ - \frac{kQ}{r} + \frac{kQ}{([L+2R]-r)}\right]_{L+R}^{R} =
( - \frac{kQ}{R} + \frac{kQ}{(L+R)}) - (-\frac{kQ}{L+R} + \frac{kQ}{(R)}) =[/tex]
[tex]\frac{2kQ}{L+R} - \frac{2kQ}{R}= \frac{2kQR-2kQ(L+R)}{R (L+R)}= \frac{2kQR-2kQL -2kQR)}{R (L+R)}= \frac{-2kQL}{R (L+R)}[/tex]

Substituting this into the formula of [tex]C= \frac{Q}{V}[/tex] provides:

[tex]C= \frac{Q}{ \frac{-2kQL}{R (L+R)}}= \frac{Q R (L+R)}{-2kQ}= \frac{R (L+R)}{-2k}[/tex]

But given is R<<L, so R= 0 which gives me a C of 0, which can't be possible! So what have I done wrong?!
 
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  • #2
Is a Taylor Expansion a good possibility?

If so this gives me:

[tex]C= \frac{-R}{2k} - \frac{-R^2}{kL} [/tex]

Is this the right way? & Have I obtained the correct answer? & How do I know if 2 terms of the expansion are enough: do I need to use only 1?
 
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  • #3
Or maybe I shouldn't let R go to zero? Then again: a negative capacitance is pretty odd...
 
  • #4
you found a negative capacitance from your negative Potential Difference.
Your Potential Difference was negative because you integrated
FROM large x-values TO small x-values, instead of along the E-field.
No big deal.

The 2kQ/R was expected ... but the Potential at the +Q was decreased
as the -Q moved in , from infinity far away to only (L+R) away.
The Potential at the -Q sphere went up as +Q came in from infinity to L+R

L >> R means that you can neglect R/L relative to 1 ... L+R = L(1 + R/L) ~ L .

careful with your bookkeeping, I think you lost an L at the end.
dV = 2kQ/R - 2kQ/L => C = R/2k + R^2/2kL
 
  • #5
So the Taylor expansion is a good idea?
 
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1. What is the 2 spheres capacitor problem?

The 2 spheres capacitor problem is a theoretical scenario in which two conducting spheres of different radii are connected by a conducting wire and separated by a dielectric material. The problem involves finding the capacitance between the spheres and the energy stored in the capacitor.

2. Why is a follow up needed on this problem?

A follow up is needed on this problem to explore any further questions or uncertainties that may have arisen during the initial analysis. It also allows for a more thorough understanding and potential solutions to the problem.

3. What are some possible solutions to the 2 spheres capacitor problem?

Some possible solutions to the 2 spheres capacitor problem include using mathematical equations and formulas to calculate the capacitance and energy stored, using simulation software to model the problem, or conducting experiments to measure the values in a real-life scenario.

4. How does this problem relate to real-world applications?

The 2 spheres capacitor problem has real-world applications in the design and functioning of electronic devices. Capacitors are used in many electronic circuits to store and regulate electrical energy, and understanding the principles behind their behavior is crucial for efficient and effective design.

5. What are some potential challenges in solving the 2 spheres capacitor problem?

Some potential challenges in solving the 2 spheres capacitor problem include dealing with non-ideal conditions, such as imperfect conductors and dielectric materials, and accounting for external factors that may affect the capacitance and energy stored, such as temperature and humidity.

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