Capacitance of the Earth

a) Discuss how the concept of capacitance can also be applied to a _single_ conductor (hint: In the relationship C = Q/V, think of the second conductor as being located at infinity).

I'm not really sure how to approach this. If I say that each conductor is infitely far apart then I can regard them both as point charges. In which case:

 V = lim r->inf [ k * q_1 * q_2 / r ] = 0

So I'm not understanding how to apply this hint.


b) Use C=Q/V to show that C = 4pi(epsilon_0) * R for a solid conducting spehre of radius R.

I think I need to get part (a) first.

c) Use your result in part (b) to calculate the capacitance of the earth, which is a good conductor of radius 6380km, compare to typical capacitors used in electronic circuits that have Capacitances from 10 pico farad to 100 micro farad




I just need help getting started.
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BrianGEFF719Asked:
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ozoCommented:
What's the voltage of a charge Q on a sphere of radus r?
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d-glitchCommented:
You can't start with point charges.
You have to start with a sphere.

Put a charge Q on a spere of radius r, calculate the E field (magnitude and direction), and think about where it goes to zero.

The dimensions of an E field are volts/meter, so if you integrate along a field line you will get a voltage.


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InteractiveMindCommented:
b)

Gauss's Law for E:  Integral{over S} E•dS = Sum_of Q/eps_0

thus:

E(4*PI*r²) = Q/eps_0

E = Q / (4*PI*r²*eps_0)

E = V/r, hence:

V = Q / (4*PI*r*eps_0)

C = Q/V = 4*PI*r*eps_0
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InteractiveMindCommented:
Better yet:

Coulomb's Law:  E = Q / (4*PI*r²*eps_0)

V= - Integral E dr, per se d-glitch's comment;

V = - Integral Q / (4*pi*r²*eps_0) dr

And so on
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InteractiveMindCommented:
By the way, it would seem that the permittivity of a conductor affects it's capacitance; so I think it should really be:
C=4*PI*r*eps_s
(where eps_s, the static permittivity, =eps_0*eps_r)
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aburrCommented:
When your problem suggested that you consider the other conductor to be at infinity, it was not suggesting that it be a point at infinity, but was suggesting that the other conductor be a sphere with an infinite radius. The potential of that very large sphere can be taken to be zero.
The permitivity to be used is the permitivity of the medium in which the electric field lines exist. In this case, free space. (There are no field lines inside a conductor.)
Modern capacitors can have values greater that 1 farad.
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