Given the electric flux find the total charge in a tiny sphere

[kacete]

Homework Statement

Given the Electric flux density of [tex]\vec{D}=\frac{100xy}{z^{2}+1}\vec{u_{x}}+\frac{50x^{2}}{z^{2}+1}\vec{u_{y}}+\frac{100x^{2}yz}{(z^{2}+1)^{2}}\vec{u_{x}} C/m^{2}[/tex] find the total charge inside a tiny sphere with a radius of [tex]r=1\mu m[/tex] centered in [tex]c(5,8,1)[/tex].

Homework Equations

According to Gauss' Law
[tex]\oint\vec{D}.d\vec{s}=Q_{involved charge}[/tex]
Solution
[tex]Q=2,26.10^{-14}C[/tex]

The Attempt at a Solution

In the previous exercises, the Electric flux expression was in spheric coordinates, which was easy to integrate using Gauss' Law. In this one I don't know where to begin, cause I tried converting it to spheric coordinates but it turned out to become a huge equation and there must be an easier way to solve it.

 

[gabbagabbahey]

You might try using the divergence theorem here

 

[kacete]

I got the following result:

[tex]\vec{\nabla}.\vec{D}=\frac{100y}{z^{2}+1}-\frac{400x^{2}y^{2}z}{(z^{2}+1)^{3}}[/tex]

Then I replaced with the coordinates and got:

[tex]\vec{\nabla}.\vec{D}=-79600[/tex]

Calculated the sphere volume:

[tex]V=\frac{4\pi r^{3}}{3}[/tex]

And multiplied both:

[tex]Q=-3.334.10^{-13}[/tex]

It's close, but still not the right solution, what did I do wrong?

 

[gabbagabbahey]

I got the following result:

[tex]\vec{\nabla}.\vec{D}=\frac{100y}{z^{2}+1}-\frac{400x^{2}y^{2}z}{(z^{2}+1)^{3}}[/tex]

You seem to be missing a term here.

Then I replaced with the coordinates and got:

[tex]\vec{\nabla}.\vec{D}=-79600[/tex]

You replaced with what coordinates, and why?

 

[kacete]

You seem to be missing a term here.

No, since it does not contain an y variable, the derivate in order of y is 0.

You replaced with what coordinates, and why?

I replaced the variables with the coordinates for the center of the sphere to find the value in that region.

 

[gabbagabbahey]

No, since it does not contain an y variable, the derivate in order of y is 0.

You are still missing a term:

[tex]\frac{\partial}{\partial z}\left(\frac{100x^2 y z}{(z^2+1)^2}\right)\neq -\frac{400x^{2}y^{2}z}{(z^{2}+1)^{3}}[/tex]

I replaced the variables with the coordinates for the center of the sphere to find the value in that region.

But that only gives you the value of the divergence at the center...the divergence theorem tells you to integrate the divergence over the entire volume of the sphere...which means you need to know its value everywhere in the sphere, not just at the center.

 

[kacete]

[tex]\frac{\partial}{\partial z}\left( - \frac{100x^2 y z}{(z^2+1)^2}\right)=\frac{100(3z^{2}-1)x^{2}y}{(z^{2}+1)^{3}}[/tex]
You're right, I corrected it. (plus i had a typo in my problem, the last one - in direction of z - is negative).

I thought that too. But I don't know how to integrate to the whole sphere using cartesian coordinates.

 

[gabbagabbahey]

Then switch to spherical coordinates, centered at [itex]c[/itex]...

[tex](x-5)=r\sin\theta\cos\phi[/tex]
[tex](y-8)=r\sin\theta\sin\phi[/tex]
[tex](z-1)=r\cos\theta[/tex]

...

 

[kacete]

Of course, but I have to switch the divergence calculated before to spherical coordinates too, right?

 

[gabbagabbahey]

Yes, of course...

 

[kacete]

Is there no easier way of solving this rather than integrating the following?

[tex]
\int^{10^{-6}}_{0} \int^{ \pi }_{0} \int^{2 \pi }_{0} \left( \frac{100(rsin \theta sin \phi +8)}{(rcos \ntheta +1)^{2} +1} - \frac{100(3(rcos \theta +1)^{2}-1)(rsin \theta cos \phi +5)^{2}(r sin \theta sin \phi )^{2}}{((rcos \theta +1)^{2}+1)^{3}}\right) r^{2} sin \theta d \phi d \theta dr
[/tex]

I'm having a hard time doing so. Even in my TI-89.

 

[gabbagabbahey]

First, it looks like you have some typos/minor errors in that expression that you'll want to correct.

Second, I don't think you really need an exact solution here... assuming the "1" in your [itex]z^2+1[/itex] terms is [itex]1\text{m}^2[/itex], then [itex]r^2\cos\theta\leq 10^{-12}\text{m}^2[/itex] is going to be very small in comparison and can be neglected...similar arguments can be used for [itex]r\sin\theta\sin\phi+8[/itex] and similar terms, giving you a much easier (approximate) expression to integrate.

 

[kacete]

Thank you, I will try that!