Solved

# Capacitance of the Earth

Posted on 2007-04-05
5,013 Views
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.
0
Question by:BrianGEFF719
[X]
###### Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

• Help others & share knowledge
• Earn cash & points

LVL 84

Accepted Solution

ozo earned 150 total points
ID: 18862599
What's the voltage of a charge Q on a sphere of radus r?
0

LVL 27

Assisted Solution

d-glitch earned 100 total points
ID: 18863368

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.

0

LVL 25

Assisted Solution

InteractiveMind earned 150 total points
ID: 18863379
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
0

LVL 25

Assisted Solution

InteractiveMind earned 150 total points
ID: 18863519
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
0

LVL 25

Expert Comment

ID: 18864164
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)
0

LVL 27

Assisted Solution

aburr earned 100 total points
ID: 18864680
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.
0

## Featured Post

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

### Suggested Solutions

Group Data Frequency Distribution 9 53
Error in calculation 3 69
How to install SVN Command Line Client? 5 133
Just learning algebra / need solution explained 4 82
Lithium-ion batteries area cornerstone of today's portable electronic devices, and even though they are relied upon heavily, their chemistry and origin are not of common knowledge. This article is about a device on which every smartphone, laptop, an…
This article provides a brief introduction to tissue engineering, the process by which organs can be grown artificially. It covers the problems with organ transplants, the tissue engineering process, and the current successes and problems of the tec…
This is a video describing the growing solar energy use in Utah. This is a topic that greatly interests me and so I decided to produce a video about it.
Finds all prime numbers in a range requested and places them in a public primes() array. I've demostrated a template size of 30 (2 * 3 * 5) but larger templates can be built such 210  (2 * 3 * 5 * 7) or 2310  (2 * 3 * 5 * 7 * 11). The larger templa…
###### Suggested Courses
Course of the Month4 days, 18 hours left to enroll