When is dA not integrated into A for Gauss' law?

In summary: If your area is not symmetric or if the E field is not constant over the area, then you'll have to do some more work to integrate the law.
  • #1
glennpagano
17
0
I have been studying Gauss' law and almost all of the problems I have been doing just have me integrate dA alone into A. I was wondering when do you actually have to do some more in depth integration.
 
Physics news on Phys.org
  • #2
The problems are usually set up with some nice symmetry so you can do the integral trivially. If your area was some crazy shape, or if your b field was not constant then you would actually have to integrate.
 
  • #3
In general,

[tex]\int_{\mathcal{S}}\textbf{E}\cdot d\textbf{a}=|\textbf{E}|A[/tex]

is only true when [itex]\textbf{E}[/itex] is uniform (constant) over the surface [itex]\mathcal{S}[/itex], and normal (orthogonal/perpendicular) to [itex]\mathcal{S}[/itex] at every point on the surface.
 
  • #4
What if it is not perpendicular? Then you will have E[tex]\oint cos \phi dA[/tex]. Then you can just pull out the cosine then because it is constant when compared to dA. If the E is not constant then will you still have to integrate it? Do you have to make it with respect to dA first? The more info the better. Right now, for most occasions I see Gauss' law as EA=q/epsilon naught.
 
  • #5
glennpagano said:
What if it is not perpendicular? Then you will have E[tex]\oint cos \phi dA[/tex]. Then you can just pull out the cosine then because it is constant when compared to dA.

That isn't nessecarily true. As a simple counterexample, consider a uniform E-field in the z-direction, [itex]\textbf{E}=E_0\mathbf{\hat{z}}[/itex] and integrate over a spherical surface of radius [itex]R[/itex].

[tex]\begin{aligned}\int_{\mathcal{S}}\textbf{E}\cdot d\textbf{a} &= E_0\int_0^{\pi}\int_0^{2\pi}\mathbf{\hat{z}}\cdot (R^2\sin\theta d\theta d\phi \mathbf{\hat{r}}) \\ &=E_0\int_0^{\pi}\int_0^{2\pi}R^2\sin\theta \cos\theta d\theta d\phi \\ & = 0 \\ &\neq 4\pi R^2E_0\cos\theta\end{aligned}[/tex]

If the E is not constant then will you still have to integrate it? Do you have to make it with respect to dA first? The more info the better. Right now, for most occasions I see Gauss' law as EA=q/epsilon naught.

Gauss' Law (in integral form) is really only useful (for finding [itex]\textbf{E}[/itex]) in those special cases where, due to symmetry, you can find a surface where [itex]\textbf{E}[/itex] is uniform over the surface and orthogonal to it at every point.
 

1. What is Gauss' law?

Gauss' law is a fundamental law in physics that describes the relationship between the electric flux through a closed surface and the electric charge enclosed by that surface.

2. How is Gauss' law related to dA and A?

Gauss' law involves the calculation of the electric flux, which is represented by the integral of the dot product between the electric field and the surface area element (dA). The surface area (A) is used to determine the direction and magnitude of the electric field at a point.

3. When is dA not integrated into A for Gauss' law?

Different situations may arise where dA cannot be integrated into A for Gauss' law. This includes situations where the electric field is not constant or if the surface is not a perfect shape, such as a sphere or cylinder.

4. How does not integrating dA into A affect the calculation of electric flux?

When dA cannot be integrated into A, the calculation of electric flux becomes more complicated. In these cases, the integral must be evaluated over the entire surface, rather than a simplified closed surface. This often involves breaking the surface into smaller, more manageable pieces.

5. Are there any other factors that can affect the integration of dA into A for Gauss' law?

Yes, there are other factors that can affect the integration of dA into A for Gauss' law. These can include the presence of other charges outside of the closed surface, or if the surface is not perfectly symmetrical, making it difficult to determine the direction of the electric field at every point.

Similar threads

Replies
8
Views
759
  • Classical Physics
Replies
20
Views
1K
Replies
1
Views
659
Replies
17
Views
460
Replies
10
Views
945
  • Classical Physics
Replies
10
Views
965
Replies
7
Views
1K
  • Classical Physics
Replies
5
Views
2K
  • Classical Physics
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
550
Back
Top