- #1

Eruditee

- 13

- 0

## Homework Statement

The problem states that a cube encloses charge. This cube is given in three space by <0,0,0> and <a,a,a>. The electric field is given by:

[itex]\hat{E}=\frac{4e}{a^{2}e_{0}}[\frac{xy}{a^{2}}\hat{i}+\frac{(y-x)}{a}\hat{j}+\frac{xyz}{a^{2}}\hat{k}][/itex]. I am to find the total charge enclosed using both methods or "sides" of the divergence theorem equivalence.

## Homework Equations

[itex]\displaystyle \iiint \limits_U \left({\nabla \cdot \mathbf F} \right) d V = \iint \limits_{\partial U} \mathbf F \cdot \mathbf n \ d S[/itex]

[itex]\nabla=\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}[/itex]

## The Attempt at a Solution

The first half worked out fine and makes sense, so I think this part is right. Do note; I do get a bit sloppy with notation; this is a rough draft:

[itex]\int_{0}^{a}\int_{0}^{a}\int_{0}^{a}\nabla\bullet EdV[/itex]

[itex]\nabla\bullet\hat{E}=\frac{4e}{a^{2}e_{0}}[

\frac{y}{a^{2}}\hat{i}+\frac{1}{a}\hat{j}+\frac{xy}{a^{2}}\hat{k}][/itex]

[itex]\frac{4e}{a^{2}e_{0}}\int_{0}^{a}\int_{0}^{a}\int_{0}^{a}(\frac{y}{a^{2}}+\frac{1}{a}+\frac{xy}{a^{2}})dV[/itex]

[itex]=\frac{4e}{a^{2}e_{0}}\int(\frac{y}{a^{2}}+\frac{1}{a}+\frac{xy}{a^{2}})dx(dA)=\frac{4e}{a^{2}e_{0}}(\int(\frac{y}{a^{2}})dx+\int\frac{1}{a}dx)+\int\frac{xy}{a^{2}}dx)dA=\frac{yx}{a^{2}}+\frac{x}{a}+\frac{x^{2}y}{2a^{2}}|_{0}^{a}=\frac{y}{a}+1+\frac{y}{2}

[\itex]

[itex]=\frac{4e}{a^{2}e_{0}}\int_{0}^{a}(\frac{y}{a}+1+\frac{y}{2}dy)dz=\frac{y^{2}}{2a}+y+\frac{y^{2}}{4a}|_{0}^{a}=\frac{a}{2}+a+\frac{a}{4}

[/itex]

[itex]=\int\frac{a}{2}+a+\frac{a}{4}dz=\frac{az}{2}+az+\frac{az}{4}=\frac{a^{2}}{2}+a^{2}+\frac{a^{2}}{4}=a^{2}+\frac{a^{2}}{2}+\frac{a^{2}}{4}=a^{2}(\frac{1}{2}+\frac{1}{4})=\frac{3a^{2}}{4}*\frac{4e}{a^{2}e_{0}}=\frac{3e}{e_{0}}

[/itex]

which makes sense, as there are 3 faces in which flux can go by?

However, the surface integral is a mess:

when z=a ; Sz(surface raised from xy by a)=a

[itex]\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet-kdxdy

=-\int_{0}^{a}\frac{xyz}{a^{2}}dxdy=-z\int_{0}^{a}\frac{xy}{a^{2}}dxdy=-\int\frac{x^{2}y}{2a^{2}}dy=\frac{-x^{2}z}{2a^{2}}\int ydy=\frac{-x^{2}zy^{2}}{4a^{2}}=-\frac{a^{3}}{4}

[/itex]

likewise:

[itex]\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet jdxdy

=\frac{1}{a}\int_{0}^{a}(y-x)dxdz=\frac{1}{a}\int yx-\frac{x^{2}}{2}dz=yxz-\frac{x^{2}z}{2}=a^{2}-\frac{a^{2}}{2}

[/itex]

[itex]\int\int\hat{F}\bullet\hat{dS}=\int\int F\bullet idydz

=\int

\frac{xy}{a^{2}}dydz=\frac{x}{a^{2}}\int ydydz=\int\frac{xy^{2}}{2}dz=\frac{xy^{2}z}{a^{2}}=\frac{a^{2}}{2}

[/itex]

And the total:

[itex]a^{2}-\frac{a^{3}}{4}=\frac{4a^{2}-a^{3}}{4}=\frac{a^{2}(4-a)}{4}

[/itex]

which I know is wrong because the flux is zero at a=4.

Where am I going wrong? I haven't done surface/volume integrals in about 4-5 years, so bear with me. I'm trying to relearn through Boas.