Calculating Capacitance of Two Isolated Conducting Spheres

[Lisa...]

The question 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?

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

The electric field due both spheres is basically Coulombs law ($$E= \frac{Q}{4 \pi \epsilon_0 r^2}$$ with the only difference is for #1 Q= Q and for #2 Q= -Q. Between the two spheres the two electric fields add (if you draw a pic, the field lines of #1 move away from #1 (cause it's positive) in radial direction, and the field lines of #2 approach #2 in radial direction (cause it's negative)).
Therefore $$E total= \frac{2Q}{4 \pi \epsilon_0 r^2}$$.

Now substitute this in the integral $$V= - \int_{#1}^{#2} E dl$$ with values of:

$$E= E total= \frac{2Q}{4 \pi \epsilon_0 r^2}$$
dl= dr, with dr the projection of dl in radial direction and
#1= -0.5 l
#2= 0.5 l

(I placed the origin exactly in the middle of both spheres, so when integrating the positions of #1 (at -0.5 l)and #2 (at 0.5 l) I need to integrate between -0.5 l and 0.5 l.

This gives:

$$V= - \int_{-0.5 l}^{0.5 l} \frac{2Q}{4 \pi \epsilon_0 r^2} dr = - \frac{2Q}{4 \pi \epsilon_0} \int_{-0.5 l}^{0.5 l} \frac{1}{r^2} dr = - \frac{2Q}{4 \pi \epsilon_0} \left[ - \frac{1}{r} \right]_{-0.5 l}^{0.5 l} = - \frac{2Q}{4 \pi \epsilon_0} \frac{-4}{l}= \frac{8Q}{4 \pi \epsilon_0 l} = \frac{2Q}{\pi \epsilon_0 l}$$

Substituting this into the formula of $$C= \frac{Q}{V}$$ provides:

$$C= \frac{Q}{\frac{2Q}{\pi \epsilon_0 l}}= \frac{Q \pi \epsilon_0 l }{2Q}= \frac{\pi \epsilon_0 l }{2}$$


So I am I right or not ...?

 

[lightgrav]

In the formula for E-field, r is the distance from the source charge to the location you're integrating. If you look at your integrand, you'll see that you went through r=0, in the middle of the integration range. But you should've started close (only R away) to the left charge, gone farther away from it, and ended close to the right charge (only R away from it).

The distance from the left sphere is r_l = x + L, (with x_minimum = L - R)
... what's the distance from the right sphere center?

 

[Lisa...]

I've edited my answer, and made the following drawing, in order to clear things up.

http://img147.imageshack.us/img147/7725/naamloos6ha.gif

Please tell me if I've understood you well:

The question 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?

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

The electric field due both spheres is basically Coulombs law ($$E= \frac{Q}{4 \pi \epsilon_0 r^2}$$ with the only difference is for #1 Q= Q and for #2 Q= -Q. Between the two spheres the two electric fields add (if you draw a pic, the field lines of #1 move away from #1 (cause it's positive) in radial direction, and the field lines of #2 approach #2 in radial direction (cause it's negative)).
Therefore $$E total= \frac{2Q}{4 \pi \epsilon_0 r^2}$$.

Now substitute this in the integral $$V= - \int_{#1}^{#2} E dl$$ with values of:

$$E= E total= \frac{2Q}{4 \pi \epsilon_0 r^2}$$
dl= dr, with dr the projection of dl in radial direction and
r is varying from a to b, because we're measuring the potential between the two spheres.
#1= point a= R (measured from the sphere 1 as starting point)
#2= point b= L-R (measured from the sphere 1 as starting point)

This gives:

$$V= - \int_{R}^{L-R} \frac{2Q}{4 \pi \epsilon_0 r^2} dr = - \frac{Q}{2 \pi \epsilon_0} \int_{R}^{L-R} \frac{1}{r^2} dr = - \frac{Q}{2 \pi \epsilon_0} \left[ - \frac{1}{r} \right]_{R}^{L-R}$$

$$V= - \frac{Q}{2 \pi \epsilon_0} (\frac{-1}{L-R} + \frac{1}{R})= = - \frac{Q}{2 \pi \epsilon_0} (\frac{-R}{R (L-R)} + \frac{(L-R)}{R (L-R)})= - \frac{Q}{2 \pi \epsilon_0} (\frac{L-2R}{R (L-R)})= \frac{-QL + 2 QR}{2 \pi \epsilon_0 R (L-R)}$$

Substituting this into the formula of $$C= \frac{Q}{V}$$ provides:

$$C= \frac{Q}{\frac{-QL + 2 QR}{2 \pi \epsilon_0 R (L-R)}}= \frac{Q 2 \pi \epsilon_0 R (L-R) }{-QL + 2 QR}= \frac{2 \pi \epsilon_0 R (L-R) }{2R - L}$$


Though this is a less compact & nice formula compared to the first one

 

[Lisa...]

I really hope somebody would help me, because I'm stuck with my work right now... thanks for your efforts!

 

[topsquark]

I really hope somebody would help me, because I'm stuck with my work right now... thanks for your efforts!

Two comments. And first I would like to say that my EM is pretty rusty, so don't take the second one as gospel...

1) You said that R<<L. If you take this limit in your final formula the capacitance becomes negative. Not a good sign.

2) I don't like your expression for the electric field. I would think it would be of the form: $$E = \frac{Q}{4 \pi \epsilon_0 r^2} - \frac{-Q}{4 \pi \epsilon_0 ([L+2R]-r)^2}$$. This reflects that the electric field of a point charge is central on the point charge. Your's is a field due to a 2Q charge.

-Dan

 

[lightgrav]

If you don't like TopSquark's origin at center of left sphere, put the origin back in the middle, (R + L/2) from each center-of-charge.
The distance from the left center-of-charge to "x" is now x + R + L/2 .
Can't you figure out what the distance from the right center-of-charge is?
Did you draw the "arbitrary integration point "x" and the increment "dx"?

To get the potential difference by integrating the E-field, choose a location "x" for an incremental path length "dx" to be drawn. draw it.
Identify the left source charge Q ; the label the distance from the center of that charge to "x". Notice that the E-field only has an x-component.
Distance from the right source charge center to point "x" is d= L/2 + R - x .
You want to integrate from x = -L/2 , to x = L/2 . . . Don't ignore the R relative to the L/2 until AFTER you've integrated, or you'll end up with 1/0 !

By the way, the above is NOT how I would answer the original Question.

You know that the Electric Potential of an isolated sphere with charge Q is
V = kQ/R . Almost all the Electric Potential accumulates very near the sphere , since E-field dies out as 1/r^2 (that is, nearby!).

So V of the positively-charged sphere is V+ = +kQ/R , while V_ = - kQ/R .
What's the Potential Difference (voltage) from one to another?

What's the capacitance?

 

[Lisa...]

By the way, the above is NOT how I would answer the original Question.

You know that the Electric Potential of an isolated sphere with charge Q is
V = kQ/R . Almost all the Electric Potential accumulates very near the sphere , since E-field dies out as 1/r^2 (that is, nearby!).

So V of the positively-charged sphere is V+ = +kQ/R , while V_ = - kQ/R .
What's the Potential Difference (voltage) from one to another?

What's the capacitance?

OMG That would indeed make the question a lot easier!

C= Q/V

and V= kQ/R - - kQ/R = 2kQ/R

Therefore C= Q/(2kQ/R) and C= QR/2kQ = R/2k...

This is what you mean right?
Thanks for helping !

 

[lightgrav]

Yeah, that's how Gene Mosca wanted you to do it.

But you should be able to calculate an E-field between 2 charges!
Stay on if you want to see that the two ways get the sanme answer.

 

[Lisa...]

LOL how did you know this is a question from Tipler and Moscas book :P? Huge & heavy book too, I must say ... ;-)

 

[lightgrav]

If we carry the extended version to every class for three semesters,
we get a gym credit for it (;>). Only 3 of us made it past Halloween ...

Do you see how you would otherwise add the E-field vector contributions?
(you know you'll get a chance to integrate an E-field on the exam ...)

 

[Lisa...]

To be honest, I really don't know how on Earth I will integrate:

$$\int_{-0.5 L}^{0.5 L} \frac{Qk}{(R+x+0.5L)^2} + \frac{Qk}{(R-x+0.5L) ^2} dx$$

The biggest problem is that we're having EM courses, without having a few math courses first (in order to learn tricks for integration / work with spherical coordinates & double integrals...)

 

[lightgrav]

R and L are just constants ... so are k and Q.
if the thing in parentheses is distance "t" = R+x+L/2, then dx = dt ,
and you only have to do : kQ int [dt/t^2] = kQ int [ t^-2 dt] = - kQ/t.

Of course the other term has an extra negative to watch out for ...
You ARE doing plain integrations, right?

 

[Lisa...]

Yeah of course I am... :P I apologise for not seeing that R & L are constants, I feel so stupid :P... guess I'm still sleepy ;-)
But I've gotten to the correct answer by now... :)
Thanks for your explanations & most of all for your patience!

 

[lightgrav]

me too, I'd better get to bed before the sun comes up.