Find Flux Through Cube & Sphere

[dpa]

Flux through sphere

Homework Statement

Given [itex]\vec{F}=\frac{\vec{r}}{r^2} [/itex] and unit sphere, find the flux through the surface of the cube.

Homework Equations

Surface Integral of F dS=volume integral of Div. F d^3r

The Attempt at a Solution

After the above formula, I do not have idea how to use divergence in spherical coordinate system.

 

[BruceW]

why not do it the easier way by using F dS ?

 

[CAF123]

I do not have idea how to use divergence in spherical coordinate system.

You do not need to worry so much about the divergence in SPs. Just compute the divergence of ##\vec{F}## using the identity for ##\nabla \cdot (\phi \vec{a})##, where ##\phi## and ##\vec{a}## are scalar and vector fields respectively.

 

[vanhees71]

First of all you have to specify the cube, i.e., its location and orientation. If the origin is contained inside the cube, it's not a good idea to use Gauss's Law and the volume integral over the divergence, because you have to find out how to treat the non-trivial singularity at the origin.

Last but not least, it's a pretty unusual long-ranged field. Are you sure that there isn't [itex]r^3[/itex] in the denominator? Better check your problem again!

 

[Nickie]

I would suggest using the Divergence Theorem to solve this problem. The Divergence Theorem states that the surface integral of a vector field over a closed surface is equal to the volume integral of the divergence of that vector field over the enclosed volume. In this case, the enclosed volume would be the cube.

To use the Divergence Theorem, we first need to find the divergence of the vector field \vec{F}. In spherical coordinates, the divergence can be expressed as:

div \vec{F} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 F_r) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (\sin \theta F_\theta) + \frac{1}{r \sin \theta} \frac{\partial F_\phi}{\partial \phi}

Where F_r, F_\theta, and F_\phi are the radial, polar, and azimuthal components of the vector field, respectively.

Using the given vector field \vec{F}=\frac{\vec{r}}{r^2}, we can find its divergence to be:

div \vec{F} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \frac{r}{r^2}) = \frac{2}{r}

Now, we can use the Divergence Theorem to relate the surface integral of \vec{F} over the unit sphere to the volume integral of the divergence over the enclosed volume (the cube):

\int_S \vec{F} \cdot d\vec{S} = \iiint_V div \vec{F} dV

Since the unit sphere has a radius of 1, the surface integral can be simplified to:

\int_S \vec{F} \cdot d\vec{S} = \int_S \frac{\vec{r}}{r^2} \cdot d\vec{S} = \int_S \frac{\vec{r}}{1} \cdot d\vec{S} = \int_S \vec{r} \cdot d\vec{S}

Now, we can use the fact that the surface integral of a vector field over a closed surface is equal to the volume integral of the divergence over the enclosed volume. In