# Electrostatic energy density

1. Nov 22, 2008

### abcdefg10645

The original web

http://farside.ph.utexas.edu/teaching/em/lectures/node56.html

Reading the web which is posted above

I got a question ....

I need some masters to help me!

My question :

why not to remain the left term of eq539 instead of eliminating it by

"Let us assume that is a sphere, centred on the origin, and let us take the limit in which the radius of this sphere goes to infinity. We know that, in general, the electric field at large distances from a bounded charge distribution looks like the field of a point charge, and, therefore, falls off like . Likewise, the potential falls off like . However, the surface area of the sphere increases like . Hence, it is clear that, in the limit as , the surface integral in Eq. (593) falls off like , and is consequently zero." (the photograph can be seen in that web )

Last edited: Nov 22, 2008
2. Nov 22, 2008

### Hootenanny

Staff Emeritus
Welcome to Physics Forums.

Lets look at each of the terms individually. First we have the electric potential $\phi$. Now, we are considering the integral over the entire space, that is when the radius of the sphere of integration tends to infinity. When we are a large distance from any charge distribution, this distribution may be approximated as a point charge, and as you well know the potential of a point charge is inversely proportional to the distance from the point charge. Therefore, at large distances the electric potential may be approximated thus

$$\phi\approx \frac{1}{r}$$

Similarly for the electric field, when we are a large distance from any charge distribution, this distribution appears point like (imagine looking at a golf ball from a kilometre away). Now, the electric field of a point charge is inversely proportional to the square of the distance from the point charge. Therefore, at large distances (i.e. as the radius tends to infinity) the electric field may be approximated as

$$E \approx \frac{1}{r^2}$$

Therefore, the entire integrand may be approximated as

$$\phi E \approx \frac{1}{r}\frac{1}{r^2} = \frac{1}{r^3}$$

Now, we are integrating over the surface of the sphere, which is proportional to the square of the radius,

$$S \approx r^2$$

Now taking the limit as $r\to\infty$ you should be able to see that the integrand ($\phi E \approx 1/r^3$) converges to zero faster than the measure of the domain of integration (surface area of the sphere) diverges. Therefore, the entire integral converges to zero.

Hopefully this makes intuitive sense, but it can also be proven rigorously using the mean value theorem.

3. Nov 22, 2008

### abcdefg10645

Thanks you, MASTER !

And I have other questions , the first is :

E~1/r^2 >>>>> E^2~1/r^4

and dV~r^3

multiplying together >>>> (EdV)~(1/r)

but there comes a issue!

As r approaches infinite

EdV should be canceled! (using the same conclusion you mentioned above)

!!!!Why (integral Edv) is still remaining?

the second is:

As you mentioned above(the last sentance)

"it can also be proven rigorously using the mean value theorem"

I don't know that before, and it spured my curiosity!

Can you tell me how to prove it by using "mean value theorem"

Thank you again, master!

Last edited: Nov 22, 2008
4. Nov 23, 2008

### Hootenanny

Staff Emeritus
There's no need to call me master, in fact, its just a little bit creepy...
Ahh, but the argument I made above was for a surface, not a volume. The argument is not valid for volume integration.

Imagine that we have a huge glass sphere of radius 1km and at the centre of this sphere is suspended a golf ball. Now imagine that you are stood on the surface of the sphere looking at the gold ball, no matter where you stand on the surface of that sphere the golf ball will always look like a point. Next imagine that you are inside the sphere and have a jet pack so that you can travel anywhere you like inside the volume of sphere. Now, if you stay near the outside of the sphere of course the golf ball will still look like a point. However, if you travel close to the centre of the sphere you will see that the golf ball is actually a sphere of non-zero radius. Therefore, the closer you travel towards the centre of the sphere, the less accurate the approximation that the golf ball is a point becomes.

This is analogous to the difference between the volume and surface integral in this case. Everywhere on the surface of integration the charge distribution will appear point like. However, this is not the case of the volume integral since there are points in the volume of the sphere of integration where the charge distribution does not appear point like.

Does that make sense?
The general outline of the proof is as follows:

The mean value theorem for a function $f\in C\left(\bar{\Omega}\right)$ (a function that is smooth in the closure of the domain $\Omega$

$$\exists \; \bold{x}\in\Omega \; : \;\int_\Omega f\left(\bold{t}\right)d\bold{t} = f\left(\bold{x}\right)\mu\left(\Omega\right)$$

Which in words means that there exists a point in omega such that the integral of a f over omega is equal to the value of f at x multiplied by the measure of omega.

Now, we are integrating over the surface of a sphere and our integrand is

$$f\left(\bold{t}\right) = \phi E \approx = \frac{1}{\bold{t}^3}$$

Notice that the integrand only depends on the distance from the origin (i.e. the radius of the sphere), which means that the integrand (f) takes the same value at any point on the surface of a sphere. Therefore, if we are integrating over the surface of a sphere, then the integrand (f) takes the same value at any point on the domain of integration.

Now, the mean value theorem states that there must exists a point in the domain of integration such that the integral of f over the domain is equal to the value of f at this point multiplied by the measure of the domain. However, as we have just seen in the previous paragraph f takes the same value at every point on the domain of integration. Therefore, the integral of f over the domain must always be equal to to the value of f at this point multiplied by the measure of the domain. Hence, we may write

$$\int_S \frac{1}{\bold{t}^3}d\bold{t} = \frac{1}{\bold{t}^3}\mu\left(\S\right)$$

Where S is the surface of a sphere. Hence,

$$\int_S \phi\bold{E}\cdot d\bold{S} = \frac{4\pi R^2}{R^3} = \frac{4\pi}{R}$$

And taking the limit

$$\lim_{R\to\infty} \int_S \phi\bold{E}d\bold{S} = \lim_{R\to\infty}\frac{4\pi}{R} = 0$$

Do you follow?

5. Nov 23, 2008

### abcdefg10645

Of course! Your perfect analogy does remove the abstraction of the integral!
The general outline of the proof is as follows:

The mean value theorem for a function $f\in C\left(\bar{\Omega}\right)$ (a function that is smooth in the closure of the domain $\Omega$

$$\exists \; \bold{x}\in\Omega \; : \;\int_\Omega f\left(\bold{t}\right)d\bold{t} = f\left(\bold{x}\right)\mu\left(\Omega\right)$$

Which in words means that there exists a point in omega such that the integral of a f over omega is equal to the value of f at x multiplied by the measure of omega.

Now, we are integrating over the surface of a sphere and our integrand is

$$f\left(\bold{t}\right) = \phi E \approx = \frac{1}{\bold{t}^3}$$

Notice that the integrand only depends on the distance from the origin (i.e. the radius of the sphere), which means that the integrand (f) takes the same value at any point on the surface of a sphere. Therefore, if we are integrating over the surface of a sphere, then the integrand (f) takes the same value at any point on the domain of integration.

Now, the mean value theorem states that there must exists a point in the domain of integration such that the integral of f over the domain is equal to the value of f at this point multiplied by the measure of the domain. However, as we have just seen in the previous paragraph f takes the same value at every point on the domain of integration. Therefore, the integral of f over the domain must always be equal to to the value of f at this point multiplied by the measure of the domain. Hence, we may write

$$\int_S \frac{1}{\bold{t}^3}d\bold{t} = \frac{1}{\bold{t}^3}\mu\left(\S\right)$$

Where S is the surface of a sphere. Hence,

$$\int_S \phi\bold{E}\cdot d\bold{S} = \frac{4\pi R^2}{R^3} = \frac{4\pi}{R}$$

And taking the limit

$$\lim_{R\to\infty} \int_S \phi\bold{E}d\bold{S} = \lim_{R\to\infty}\frac{4\pi}{R} = 0$$

Do you follow?[/QUOTE]

I got it ! Your interpretations do help me a lot !

THX again and again!

MR.H (May I call you like this?)

Last edited: Nov 23, 2008
6. Nov 23, 2008

### Hootenanny

Staff Emeritus
My pleasure.
:rofl: You can if you like, but everyone else calls me Hoot.

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