A tricky problem on capacity

  • Thread starter sagardip
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Homework Statement



Two identical conducting spheres of radii "a" are kept in contact. Find the capacity of the system

Homework Equations



C=Q/V

The Attempt at a Solution



Since the two spheres are kept in contact so they are at the same potential.So we need to find out this potential so as to find the capacitance. But I am not able to start at the solution
 
Last edited:

Answers and Replies

  • #2
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How far are they?
 
  • #3
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They are kept in contact
 
  • #4
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So they are placed next to each other, not through any wire?
 
  • #5
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No, directly in contact
 
  • #6
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Capacitors works creating an electric field across a dielectric, I'm pretty sure that if the conducting spheres are in touch the capacitance would be zero.
 
  • #7
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This is a very hard problem.

EDIT:

Let's take the z-axis along the line connecting the two centers of the two spheres, and the point where the spheres touch eachother lie in the xy-plane (it follows that this point coincides with the origin).

From the geometry of the problem, it follows then that the z-axis is an axis of symmetry and that the xy-plane is a plane of symmetry for the problem. Therefore, the potential [itex]\Phi[/itex] does not depend on the azimuthal angle [itex]\phi[/itex] and is also an even function of [itex]z[/itex]. It also satisfies the Laplace equation [itex]\nabla^{2} \, \Phi = 0[/itex] everywhere outside the spheres with the boundary conditions that [itex]\Phi \rightarrow 0[/itex] at infinity and [itex]\Phi = \Phi_{0}[/itex] on the spheres.

Let us try and solve this boundary value problem. Before we go any further, let us try and "invent" a new coordinate system, where one set of coordinate surfaces are spheres with the center on the z-axis and such that the origin always lies on the sphere. If the sphere has radius [itex]a[/itex], then, the equation for such a sphere in cylindrical coordinates is:

[tex]
\rho^{2} + (z - a)^{2} = a^{2}
[/tex]

[tex]
\rho^{2} + z^{2} - 2 a z = 0
[/tex]

The differential equation for this family of surfaces is obtained by eliminating the parameter [itex]a[/itex]:

[tex]
2 \rho + 2 z z' - 2 a z' = 0 \Rightarrow 2 z (\rho + z z') - 2 a z z' = 0
[/tex]

[tex]
z' (\rho^{2} + z^{2}) - 2 z (\rho + z z') = 0
[/tex]

[tex]
(\rho^{2} - z^{2}) z' - 2 \rho z = 0
[/tex]

A family of orthogonal surfaces is obtaned by the substitution [itex]z \rightarrow -1/z[/itex]:

[tex]
(\rho^{2} - z^{2}) (-\frac{1}{z'}) - 2 \rho z = 0
[/tex]

[tex]
\rho^{2} - z^{2} + 2 \rho z z' = 0
[/tex]

[tex]
z' = \frac{z^{2} - \rho^{2}}{2 \rho z}
[/tex]

This is a homogeneous differential equation and the variables can be separated by applying the substitution [itex]u = z/\rho \Rightarrow z = \rho u, z' = u + \rho u'[/itex]:

[tex]
\rho u' + u = \frac{u^{2} - 1}{2 u}
[/tex]

[tex]
\rho u' = - \frac{u^{2} + 1}{2 u}
[/tex]

[tex]
\frac{2 u du}{u^{2} + 1} + \frac{d\rho}{\rho} = 0
[/tex]

[tex]
\ln{(u^{2} + 1)} + \ln{\rho} = c'
[/tex]

[tex]
\rho (u^{2} + 1) = C
[/tex]

[tex]
\rho^{2} + z^{2} - C \rho = 0
[/tex]

But, this determines a sphere of radius [itex]|b|, \; b = C/2[/itex] lying on the [itex]\rho[/itex]-axis and touching the origin.

Let us use a and b as new coordinates instead of [itex]\rho[/itex] and z. We need to solve the system of equations:

[tex]
\left\{\begin{array}{lcl}
\rho^{2} + z^{2} - 2 a z & = & 0 \\

\rho^{2} + z^{2} - 2 b \rho & = & 0
\end{array}\right. \Rightarrow \left\{\begin{array}{l}
\rho = \frac{2 a^{2} b}{a^{2} + b^{2}} \\

z = \frac{2 a b^{2}}{a^{2} + b^{2}}
\end{array}\right. \Rightarrow r^{2} = \rho^{2} + z^{2} = \frac{4 a^{2} b^{2}}{a^{2} + b^{2}}
[/tex]

With these definitions, the metric becomes:

[tex]
ds^{2} = d\rho^{2} + dz^{2} + \rho^{2} d\phi^{2} = \frac{4(b^{4} \, da^{2} + a^{4} \, db^{2}}{(a^{2} + b^{2})^{2}} + \frac{4 a^{4} \, b^{2}}{(a^{2} + b^{2})^{2}} d\phi^{2}
[/tex]

and the Laplace operator takes the form:

[tex]
\nabla^{2} \Phi = \frac{
(a^{2} + b^{2})^{3}}{4 b^{4}} \, \frac{\partial}{\partial a}\left(\frac{a^{4}}{a^{2} + b^{2}} \, \frac{\partial \Phi}{\partial a}\right) + \frac{(a^{2} + b^{2})^{3}}{4 a^{2} \, b^{3}} \, \frac{\partial}{\partial b}\left(\frac{b}{a^{2} + b^{2}} \, \frac{\partial \Phi}{\partial b}\right) + \frac{(a^{2} + b^{2})^{2}}{4 a^{4} b^{2}} \, \frac{\partial^{2} \Phi}{\partial \phi^{2}} = 0
[/tex]

Assuming azimuthal symmetry, this potential does not depend on [itex]\phi[/itex]. However, the equation does not allow separation of the variables a and b.
 
Last edited:
  • #8
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0
when the two spheres are kept in contact, the capacitance of the system of two spheres will be zero..
but mind you, its not that what you are being asked. Its capacity of the spheres!!
for example, when we are asked to determine the capacity of a sphere carrying a charge, we assume the other concentric sphere is at infinity with opposite charge, so we can find the expression for electric field and hence integrate E.ds with apt limit to find the potential difference and hence the capacity of the sphere,
the question here is how can you determine one unique expression of electric field in this case at any point in space???
 
  • #9
394
0
edit: i was wrong
 
Last edited:
  • #10
gabbagabbahey
Homework Helper
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As you know, the voltage at the surface of a sphere that has charge q on it is the same as the voltage at that same radius of a point charge q.

We can then say:
[tex]V_{sphere}=\frac{kQ}{R}[/tex]
where Q is some charge on the sphere and r is its radius.
[tex]C =\frac{Q}{V}= \frac{Q}{\frac{kQ}{R}}=\frac{R}{k}=4\pi\epsilon_oR[/tex]

This is misleading, a single conducting spherical shell does not qualify as a capacitor. Moreover, the [itex]V[/itex] and [itex]Q[/itex] in the formula [itex]C=\frac{Q}{V}[/itex] are defined as the potential difference between the two plates of the capacitor and the charge on the positive plate respectively (the other plate must have a charge of [itex]-Q[/itex]).

When you have two spheres separated by a, it's like two point charges of Q/2 separated by 2a where Q is the total charge in the system.

So since voltage is linear, you can add the capacitance of each separate sphere to find the capacitance of the two spheres together.
[tex]C_{1sphere} =\frac{\frac{Q}{2}}{V}= \frac{\frac{Q}{2}}{\frac{k\frac{Q}{2}}{R}}=4\pi\epsilon_oR[/tex]
so
[tex]C_{2spheres}=2C_{1sphere}=8\pi\epsilon_oR[/tex]

This is just wrong. According to your calculations, the Capacitance of two idendical conductiing spheres is the same no matter how far apart they are. Capacitance is a purely geometrical quantity, and should definately depend on how far apart the two plates (spheres in this case) are.
 
Last edited:
  • #11
gabbagabbahey
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This is a very hard problem.

EDIT:

Let's take the z-axis along the line connecting the two centers of the two spheres, and the point where the spheres touch eachother lie in the xy-plane (it follows that this point coincides with the origin).

From the geometry of the problem, it follows then that the z-axis is an axis of symmetry and that the xy-plane is a plane of symmetry for the problem. Therefore, the potential [itex]\Phi[/itex] does not depend on the azimuthal angle [itex]\phi[/itex] and is also an even function of [itex]z[/itex]. It also satisfies the Laplace equation [itex]\nabla^{2} \, \Phi = 0[/itex] everywhere outside the spheres with the boundary conditions that [itex]\Phi \rightarrow 0[/itex] at infinity and [itex]\Phi = \Phi_{0}[/itex] on the spheres.

Let us try and solve this boundary value problem. Before we go any further, let us try and "invent" a new coordinate system, where one set of coordinate surfaces are spheres with the center on the z-axis and such that the origin always lies on the sphere. If the sphere has radius [itex]a[/itex], then, the equation for such a sphere in cylindrical coordinates is:

[tex]
\rho^{2} + (z - a)^{2} = a^{2}
[/tex]

[tex]
\rho^{2} + z^{2} - 2 a z = 0
[/tex]

The differential equation for this family of surfaces is obtained by eliminating the parameter [itex]a[/itex]:

[tex]
2 \rho + 2 z z' - 2 a z' = 0 \Rightarrow 2 z (\rho + z z') - 2 a z z' = 0
[/tex]

[tex]
z' (\rho^{2} + z^{2}) - 2 z (\rho + z z') = 0
[/tex]

[tex]
(\rho^{2} - z^{2}) z' - 2 \rho z = 0
[/tex]

A family of orthogonal surfaces is obtaned by the substitution [itex]z \rightarrow -1/z[/itex]:

[tex]
(\rho^{2} - z^{2}) (-\frac{1}{z'}) - 2 \rho z = 0
[/tex]

[tex]
\rho^{2} - z^{2} + 2 \rho z z' = 0
[/tex]

[tex]
z' = \frac{z^{2} - \rho^{2}}{2 \rho z}
[/tex]

This is a homogeneous differential equation and the variables can be separated by applying the substitution [itex]u = z/\rho \Rightarrow z = \rho u, z' = u + \rho u'[/itex]:

[tex]
\rho u' + u = \frac{u^{2} - 1}{2 u}
[/tex]

[tex]
\rho u' = - \frac{u^{2} + 1}{2 u}
[/tex]

[tex]
\frac{2 u du}{u^{2} + 1} + \frac{d\rho}{\rho} = 0
[/tex]

[tex]
\ln{(u^{2} + 1)} + \ln{\rho} = c'
[/tex]

[tex]
\rho (u^{2} + 1) = C
[/tex]

[tex]
\rho^{2} + z^{2} - C \rho = 0
[/tex]

But, this determines a sphere of radius [itex]|b|, \; b = C/2[/itex] lying on the [itex]\rho[/itex]-axis and touching the origin.

Let us use a and b as new coordinates instead of [itex]\rho[/itex] and z. We need to solve the system of equations:

[tex]
\left\{\begin{array}{lcl}
\rho^{2} + z^{2} - 2 a z & = & 0 \\

\rho^{2} + z^{2} - 2 b \rho & = & 0
\end{array}\right. \Rightarrow \left\{\begin{array}{l}
\rho = \frac{2 a^{2} b}{a^{2} + b^{2}} \\

z = \frac{2 a b^{2}}{a^{2} + b^{2}}
\end{array}\right. \Rightarrow r^{2} = \rho^{2} + z^{2} = \frac{4 a^{2} b^{2}}{a^{2} + b^{2}}
[/tex]

With these definitions, the metric becomes:

[tex]
ds^{2} = d\rho^{2} + dz^{2} + \rho^{2} d\phi^{2} = \frac{4(b^{4} \, da^{2} + a^{4} \, db^{2}}{(a^{2} + b^{2})^{2}} + \frac{4 a^{4} \, b^{2}}{(a^{2} + b^{2})^{2}} d\phi^{2}
[/tex]

and the Laplace operator takes the form:

[tex]
\nabla^{2} \Phi = \frac{
(a^{2} + b^{2})^{3}}{4 b^{4}} \, \frac{\partial}{\partial a}\left(\frac{a^{4}}{a^{2} + b^{2}} \, \frac{\partial \Phi}{\partial a}\right) + \frac{(a^{2} + b^{2})^{3}}{4 a^{2} \, b^{3}} \, \frac{\partial}{\partial b}\left(\frac{b}{a^{2} + b^{2}} \, \frac{\partial \Phi}{\partial b}\right) + \frac{(a^{2} + b^{2})^{2}}{4 a^{4} b^{2}} \, \frac{\partial^{2} \Phi}{\partial \phi^{2}} = 0
[/tex]

Assuming azimuthal symmetry, this potential does not depend on [itex]\phi[/itex]. However, the equation does not allow separation of the variables a and b.

Haven't you already been warned about providing too much help for homework problems? In any case, you misinterpreted the problem.
 
  • #12
gabbagabbahey
Homework Helper
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Homework Statement



Two identical conducting spheres of radii "a" are kept in contact. Find the capacity of the system

Homework Equations



C=Q/V

The Attempt at a Solution



Since the two spheres are kept in contact so they are at the same potential.So we need to find out this potential so as to find the capacitance. But I am not able to start at the solution

If two conductors (any sphape and size) are in contact, what must be the potential difference between them (and why?:wink:)? What does that make [itex]V[/itex] in the capacitance formula [itex]C=\frac{Q}{V}[/itex]?
 
  • #13
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100
Don't forget that even a single metal has capacitance albeit very small.A simple way to solve this problem would be to consider the two spheres being separated,one of them given a charge and the two then joined together.This is how I see the problem and I think it works.
 
  • #14
vela
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This is misleading, a single conducting spherical shell does not qualify as a capacitor.
Actually, it does.

http://en.wikipedia.org/wiki/Capacitance#Self-capacitance

When I was taking E&M as an undergrad and first heard the question "What's the capacitance of a sphere?" I remember thinking, "That doesn't make sense. Where's the other conductor?" It turned out the question was about this notion of self-capacitance.
 
  • #15
vela
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Homework Statement



Two identical conducting spheres of radii "a" are kept in contact. Find the capacity of the system

Homework Equations



C=Q/V

The Attempt at a Solution



Since the two spheres are kept in contact so they are at the same potential.So we need to find out this potential so as to find the capacitance. But I am not able to start at the solution
This problem suggests to me using a conformal mapping. Is that something you've covered? Unfortunately, I don't remember how to do this type of problem anymore, but I do remember that with the right mapping, the problem typically becomes very easy to solve.

I found a good list of conformal mappings at

http://math.fullerton.edu/mathews/c2003/ConformalMapDictionary.1.html [Broken]

Map 13, in particular, might be applicable.
 
Last edited by a moderator:
  • #16
gabbagabbahey
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Actually, it does.

http://en.wikipedia.org/wiki/Capacitance#Self-capacitance

When I was taking E&M as an undergrad and first heard the question "What's the capacitance of a sphere?" I remember thinking, "That doesn't make sense. Where's the other conductor?" It turned out the question was about this notion of self-capacitance.

Thanks, I don't remember ever studying self-capacitance... with this concept it seems as though the other conductor is a theoretical spherical shell of infinite radius. This seems like a useless concept to me... whenever you add charge to a conductor, it has to come from somewhere else (charge conservation). To me, that somewhere else is always the other half of the capacitor, not some theoretical infinite radius spherical shell. I guess there must be some practical use for this concept though?
 
  • #17
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100
Actually, it does.

http://en.wikipedia.org/wiki/Capacitance#Self-capacitance

When I was taking E&M as an undergrad and first heard the question "What's the capacitance of a sphere?" I remember thinking, "That doesn't make sense. Where's the other conductor?" It turned out the question was about this notion of self-capacitance.

When we find the potential of a metal(V=Q/4pi* epsilon* r)for a sphere of radius r,V stands for the potential with reference to a zero potential in other words it stands for potential difference and the other conductor can be considered to be the earth.
 
  • #18
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100
Thanks, I don't remember ever studying self-capacitance... with this concept it seems as though the other conductor is a theoretical spherical shell of infinite radius. This seems like a useless concept to me... whenever you add charge to a conductor, it has to come from somewhere else (charge conservation). To me, that somewhere else is always the other half of the capacitor, not some theoretical infinite radius spherical shell. I guess there must be some practical use for this concept though?

I can't think of any uses but would be interested to hear if there are any.The capacitance is incredibly small being just a small fraction of a Farad for a sphere as big as the earth.
 
  • #19
gabbagabbahey
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Anyways, it definitely sounds like the problem is to calculate the self-capacitance of the two-sphere conductor. Since the problem has azimuthal symmetry (if you set up your Coordinate system so that the spheres are described by [itex]x^2+y^2+(z\pm a)^2=a^2[/itex] ), I'd use separation of variables in Cylindrical Coordiantes to solve Laplace's equation. Your boundary conditions should be obvious: the potential at infinity is zero, and the spheres are at some constant potential [itex]V_0[/itex]. Finding the potential everywhere will allow you to calculate the charge density on the conductor, and the total charge as a function of [itex]V_0[/itex]...divide that charge by [itex]V_0[/itex], and you'll be left with the self-capacitance.
 
  • #20
vela
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Thanks, I don't remember ever studying self-capacitance... with this concept it seems as though the other conductor is a theoretical spherical shell of infinite radius. This seems like a useless concept to me... whenever you add charge to a conductor, it has to come from somewhere else (charge conservation). To me, that somewhere else is always the other half of the capacitor, not some theoretical infinite radius spherical shell. I guess there must be some practical use for this concept though?
I suppose it might be useful to model a system where charge collects on a conductor. That section from Wikipedia mentioned two applications: a van de Graaff generator and the Earth.

I think it's used primarily to torture poor students like sagardip. Of course, classical electromagnetics is one of those subject that been around too long in that people have had time to solve all sorts of weird problems that now get assigned as homework. :wink:
 
  • #21
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Coulombs law gives the equation I wrote in post 17 which in turn gives the capacitance of a single sphere.If the charged sphere is connected to the uncharged sphere charge will flow(half of it) until the potentials equalise.
 
Last edited:
  • #22
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I am really thankful to you for your valuable suggestions.But since I am a student of class 12th I have not yet learnt about the use of cylindrical coordinates to solve Laplace's
equation.So it would be really kind of you if you explained it to me step by step.
Moreover the answer to the question is 8*pi*epsilon*a*ln(2).Thank you.
 
  • #23
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Dear Vela, I am really sorry that I have not yet covered the topic Conformal Mapping.So it would be really kind of you to explain it to me.Thanking you in advance.
 
  • #24
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Thank you for your suggestions. Even i m a 12th grader and stuck with the same problem, it would be really kind of you if you could provide the brief explanation of conformal mapping and using of cylindrical co-ordinates etc.
If its difficult, i would request you to give suggestions of some books, or websites from where these things could be learned!!

Thank You!
 
  • #25
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100
The solution to this seems to be harder than I originally thought and whatever the equation is I feel uneasy about it for the following reasons:
1.The equation gives what vela referred to as the "self capacitance" but this assumes that the spheres are isolated.The prescence,however,of other conductors can change the potential and therefore the capacitance.In this case,where the capacitance is very small,can the surroundings be considered to have negligible effects?
2.How easy or otherwise is it to confirm any equation by experiment.Have the experiments been done?
I don't know the answers to these questions yet or even if they are relevant but will give them some thought.
Anyway,back to the problem in hand.In post 14 velas reference pointed out that La Places equation can be used to get a solution and added that "the solution is trivial for high symmetry cases".For people struggling with this problem I suggest you look back again at velas post(14) and then check out La Places equation.I bet I wouldn't find the solution "trivial":biggrin:
 
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