abcdefg10645 said:
There's no need to call me master, in fact, its just a little bit creepy...
abcdefg10645 said:
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?
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?
abcdefg10645 said:
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!
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?
)
