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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.

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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Start your 7-day free trialYou 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.

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

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

C=4*PI*r*eps_s

(where eps_s, the static permittivity, =eps_0*eps_r)

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.

Math / Science

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