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

In summary, the electric charge inside a tiny sphere with a radius of 1\mu m centered in c(5,8,1) is 2,26.10^{-14}C.
  • #1
kacete
27
0

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.
 
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  • #2
You might try using the divergence theorem here:wink:
 
  • #3
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?
 
  • #4
kacete said:
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?
 
  • #5
gabbagabbahey said:
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.
gabbagabbahey said:
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.
 
  • #6
kacete said:
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.
 
  • #7
[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.
 
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  • #8
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]

...
 
  • #9
Of course, but I have to switch the divergence calculated before to spherical coordinates too, right?
 
  • #10
Yes, of course...
 
  • #11
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.
 
  • #12
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.
 
  • #13
Thank you, I will try that!
 

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

What is electric flux?

Electric flux is a measure of the flow of electric field through a given surface. It is a scalar quantity and is represented by the symbol Φ.

How is electric flux calculated?

Electric flux is calculated by taking the dot product of the electric field and the area vector of the surface. The formula for electric flux is Φ = E * A * cos(θ), where E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the surface.

What is the unit of electric flux?

The unit of electric flux is Nm²/C, which is equivalent to volts (V). This is because electric field has the unit of N/C and area has the unit of m², so when multiplied together, the unit becomes Nm²/C.

What is the relationship between electric flux and electric charge?

Electric flux is directly proportional to the total electric charge enclosed within a surface. This means that as the electric charge increases, the electric flux also increases, and vice versa.

How can the total charge in a tiny sphere be found using electric flux?

The total charge in a tiny sphere can be found by using the formula Q = Φ * ε₀, where Q is the total charge, Φ is the electric flux, and ε₀ is the permittivity of free space. This formula is derived from Gauss's Law, which states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space.

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