Electric Flux through a surface.

In summary, the problem is to evaluate the net electric flux out of a spherical region without any charge, with a point charge located at a distance d outside the region. The electric field of a point charge and the electric flux out of a surface are given. By applying Gauss' Law, it can be determined that the net electric flux is zero, as there is no charge enclosed by the sphere. The process of solving this problem using a surface integral is described, including the use of unit vectors and the law of cosines. The final result is that the net electric flux is zero, confirming the validity of Gauss' Law and its ability to simplify problems of this type.
  • #1
Brian-san
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Homework Statement


Consider a spherical region V of radius R without any charge and a point charge q outside the spherical region at a distance d from the center as shown in the figure. Evaluate explicitly the net electric flux out of the surface S which bounds the spherical region V.

Homework Equations


Electric Field of a Point Charge (k=1):
[tex]\vec{E}=\frac{q}{r^2}\hat{r}[/tex]

Electric Flux out of a Surface:
[tex]\Phi_{E}=\int_{S}\vec{E}\cdot d\vec{\sigma}[/tex]

The Attempt at a Solution


From Gauss' Law, I know the answer should be zero since there is no charge enclosed by the sphere. The main problem with solving this by using the surface integral instead, is that I can only place either the sphere or the charge at the origin, which yields a fairly complicated integral.

If I consider the charge at the origin, then [tex]\vec{E}=\frac{q}{r^2}\hat{r}[/tex], and [tex]d\vec{\sigma}=R^2sin\theta d\Phi d\theta\hat{R}[/tex]

If I place the sphere at the origin, then [tex]d\vec{\sigma}=R^2sin\theta d\Phi d\theta\hat{r}[/tex], but the electric field is then [tex]\vec{E}=\frac{q}{r^2}\hat{r'}[/tex]

Here, due to the spherical symmetry, the integral can be simplified to [tex]\Phi_{E}=2\pi qR^2\int^{\pi}_{0}\frac{sin\theta}{r^2}(\hat{r}\cdot \hat{r'})d\theta[/tex]

For some point on the circle (R, θ), r²=R²+d²-2Rdcosθ, so [tex]\Phi_{E}=2\pi qR^2\int^{\pi}_{0}\frac{sin\theta d\theta}{R^2²+d^2-2Rdcos\theta}(\hat{r}\cdot \hat{r'})[/tex]

The problem for me is the dot product term, since both are unit vectors, it is simply the cosine of the angle between the to vectors. However, using the law of cosines again gives
[tex]cos\theta '=\frac{R^2+r^2-d^2}{2Rr}=\frac{R-dcos\theta}{\sqrt{R^2²+d^2-2Rdcos\theta}}[/tex]

Then the integral is [tex]\Phi_{E}=2\pi qR^2\int^{\pi}_{0}\frac{sin\theta (R-dcos\theta )d\theta}{(R^2²+d^2-2Rdcos\theta)^{3/2}}[/tex]

The first integral, [tex]\int^{\pi}_{o}\frac{Rsin\theta d\theta}{(R^2²+d^2-2Rdcos\theta)^{3/2}}=\frac{-1}{d\sqrt{R^2²+d^2-2Rdcos\theta}}\right|^{\pi}_{0}=\frac{-1}{d\sqrt{R^2²+d^2-2Rdcos\pi}}+\frac{1}{d\sqrt{R^2²+d^2-2Rdcos0}}[/tex]
[tex]=\frac{-1}{d\sqrt{R^2²+d^2+2Rd}}+\frac{1}{d\sqrt{R^2²+d^2-2Rd}}=\frac{-1}{d\sqrt{(R-d)^2}}+\frac{1}{d\sqrt{(R+d)^2}}=\frac{1}{d(R+d)}+\frac{-1}{d(R-d)}=\frac{-2}{(R^2-d^2)}[/tex]

I assume the above integral is correct.

The problem comes from the second integral, with the sinθcosθdθ in the numerator, it is not of a particular form I am familiar with, and there seem to be no obvious substitutions in order to integrate it. Is this solution on the right track, or is there something I missed that may have been easier? Thanks.
 
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  • #2
Brian-san said:
Then the integral is [tex]\Phi_{E}=2\pi qR^2\int^{\pi}_{0}\frac{sin\theta (R-dcos\theta )d\theta}{(R^2²+d^2-2Rdcos\theta)^{3/2}}[/tex]

So far, so good!:approve:

The first integral, [tex]\int^{\pi}_{o}\frac{Rsin\theta d\theta}{(R^2²+d^2-2Rdcos\theta)^{3/2}}=\left.\frac{-1}{d\sqrt{R^2²+d^2-2Rdcos\theta}}\right|^{\pi}_{0}=\frac{-1}{d\sqrt{R^2²+d^2-2Rdcos\pi}}+\frac{1}{d\sqrt{R^2²+d^2-2Rdcos0}}[/tex]
[tex]=\frac{-1}{d\sqrt{R^2²+d^2+2Rd}}+\frac{1}{d\sqrt{R^2²+d^2-2Rd}}=\frac{-1}{d\sqrt{(R-d)^2}}+\frac{1}{d\sqrt{(R+d)^2}}=\frac{1}{d(R+d)}+\frac{-1}{d(R-d)}=\frac{-2}{(R^2-d^2)}[/tex]

I assume the above integral is correct.

Not quite, if the point charge is outside the sphere, then [itex]d>R[/itex] and hence

[tex]\sqrt{(R-d)^2}=d-R\neq R-d[/tex]

The problem comes from the second integral, with the sinθcosθdθ in the numerator, it is not of a particular form I am familiar with, and there seem to be no obvious substitutions in order to integrate it. Is this solution on the right track, or is there something I missed that may have been easier? Thanks.

Just make the substitution [itex]x=\cos\theta[/itex] and then integrate by parts.
 
  • #3
Adjusting for the square root mistake, then the first result will be
[tex]\frac{2R}{d(R^2-d^2)}[/tex]

For the secont integral I get
[tex]\frac{\sqrt{R^2+d^2-2Rdcos\theta}}{2R^2d^2}+\frac{R^2+d^2}{2R^2d^2\sqrt{R^2+d^2-2Rdcos\theta}}\right|^{\pi}_{0}=\frac{R^2+d^2-Rdcos\theta}{R^2d^2\sqrt{R^2+d^2-2Rdcos\theta}}\right|^{\pi}_{0}[/tex]
[tex]=\frac{R^2+d^2+Rd}{2R^2d^2\sqrt{R^2+d^2+2Rd}}-\frac{R^2+d^2-Rd}{R^2d^2\sqrt{R^2+d^2-2R}}=\frac{R^2+d^2+Rd}{2R^2d^2\sqrt{(R+d)^2}}-\frac{R^2+d^2-Rd}R^2d^2\sqrt{(R-d)^2}}=\frac{R^2+d^2+Rd}{R^2d^2(R+d)}-\frac{R^2+d^2-Rd}{R^2d^2(d-R)}[/tex]
[tex]=\frac{R^2((d-R)-(d+R))+d^2((d-R)-(d+R))+Rd((d-R)+(d+R))}{R^2d^2(d+R)(d-R)}=\frac{R^2(-2R)+d^2(-2R)+Rd(2d)}{R^2d^2(d^2-R^2)}[/tex]
[tex]=\frac{-2R^3}{R^2d^2(d^2-R^2)}=\frac{-2R}{d^2(d^2-R^2)}=\frac{2R}{d^2(R^2-d^2)}[/tex]

This term is then multiplied by -d and added to the previous getting zero. This is what I expected from Gauss Law. I really can appreciate Gauss' Law in this case, it greatly simplifies this type problem.
 
  • #4
Brian-san said:
This is what I expected from Gauss Law. I really can appreciate Gauss' Law in this case, it greatly simplifies this type problem.

Indeed!:smile:
 

1. What is electric flux through a surface?

Electric flux through a surface is the measure of the electric field passing through a given surface. It is a scalar quantity and is measured in units of volts per meter (V/m).

2. How is electric flux through a surface calculated?

The electric flux through a surface is calculated by taking the dot product of the electric field vector and the surface area vector. It is given by the equation Φ = E · A, where Φ is the electric flux, E is the electric field, and A is the surface area.

3. What is the difference between electric flux and electric field?

The electric field is a vector quantity that describes the strength and direction of the electric force at a given point. Electric flux, on the other hand, is a scalar quantity that represents the amount of electric field passing through a surface.

4. How does the shape of a surface affect electric flux?

The shape of a surface can affect the electric flux through it, as the amount of surface area that the electric field passes through can vary. A surface that is perpendicular to the electric field lines will have a greater electric flux than a surface that is parallel to the field lines.

5. What is the significance of electric flux through a closed surface?

If the electric flux through a closed surface is non-zero, it indicates the presence of an electric charge within the enclosed space. This is known as Gauss's Law and is a fundamental principle in the study of electromagnetism.

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