91影视

a) To determine the formula for the electric field of a straight charged wire, we can begin using Gauss's law:

$\begin{array}{r}\text{鈭�}\stackrel{鈫�}{E}鈰�d\stackrel{鈫�}{A}=\frac{Q_{in}}{{系}_0}\end{array}$

$\text{鈭�}\stackrel{鈫�}{E}鈰�d\stackrel{鈫�}{A}=EA$(cylindrical surface)

localid="1648841083150" $EA=\frac{Q_{in}}{{系}_0}$

$E=\frac{Q_{in}}{A{系}_0}$(show E)

$A=2r蟺l$(sphere's surface)

$Q_{in}=位l$(linear charge definition)

$E=\frac{位}{2r蟺{系}_0}$

$\stackrel{鈫�}{E}=\frac{位\widehat{r}}{2r蟺{系}_0}$(outside direction)

a)

You may have been using the term "electric flux into a zone" to describe the flow of electricity in a certain area$d\stackrel{鈫�}{A}$

$d\mathrm{桅}=\stackrel{鈫�}{E}鈰�d\stackrel{鈫�}{A}$

localid="1648909065267" $d\stackrel{鈫�}{A}=Ldy\stackrel{鈭�}{i}$(yz plane's area)

$r=\sqrt{y^2+{\left(\frac{L}{2}\right)}^2}$(distance between wire and face)

localid="1648909073230" $d\mathrm{桅}=\frac{位L(\widehat{r}鈰�\stackrel{鈭�}{i})dy}{2蟺{系}_0\sqrt{y^2+{\left(\frac{L}{2}\right)}^2}}$localid="1648909081409" $\text{(put}r\text{and}E\text{to}\text{fformulaor}d\mathrm{桅}\text{)}$

$d\mathrm{桅}=\frac{位L\cos 鈦�胃dy}{2蟺{系}_0\sqrt{y^2+{\left(\frac{L}{2}\right)}^2}}$ (scalar combination definition)

$\cos 鈦�胃=\frac{\frac{L}{2}}{\sqrt{y^2+{\left(\frac{L}{2}\right)}^2}}$ (by geometry)

$d\mathrm{桅}=\frac{位Ldy}{2蟺{系}_0\sqrt{y^2+{\left(\frac{L}{2}\right)}^2}}鈰�\frac{\frac{L}{2}}{\sqrt{y^2+{\left(\frac{L}{2}\right)}^2}}$ (put $\cos 鈦�胃$)

$d\mathrm{桅}=\frac{位L^{2}dy}{4蟺{系}_0\left(y^2+{\left(\frac{L}{2}\right)}^2\right.}$

b) Then you should integrating the formula to see the net flux:

$\mathrm{桅}={鈭�}_{鈭�\frac{L}{2}}^{\frac{L}{2}}鈥�\frac{位L^{2}dy}{4蟺{系}_0\left(y^2+{\left(\frac{L}{2}\right)}^2\right.}$

$鈭�\frac{dx}{x^2+a^2}=\frac{1}{a}\arctan 鈦�\frac{x}{a}$

$\mathrm{桅}=\frac{位L^2}{4蟺{系}_0}{\left[\frac{2}{L}\arctan 鈦�\frac{2y}{L}\right]}_{鈭�\frac{L}{2}}^{\frac{L}{2}}$

$\mathrm{桅}=\frac{位L}{2蟺{系}_0}\left(\arctan 鈦�\frac{2L}{2L}鈭�\arctan 鈦�\frac{鈭�2L}{2L}\right)$(apply boundaries)

$\mathrm{桅}=\frac{位L}{2蟺{系}_0}\left(\frac{蟺}{4}鈭�\left(鈭�\frac{蟺}{4}\right)\right)$ (calculate arc tangent )

$\mathrm{桅}=\frac{位L}{4{系}_0}$

c) The flux goes through the four sections of the cube, resulting in a total flux, that is ${\mathrm{桅}}_{net}=4\mathrm{桅}$

${\mathrm{桅}}_{net}=4\mathrm{桅}$

${\mathrm{桅}}_{net}=4\frac{位L}{4{系}_0}$ (put formula for $\mathrm{桅}$)

${\mathrm{桅}}_{net}=\frac{位L}{{系}_0}$

you receive what you want to show when you apply the sequential charge density formula:

$位=\frac{Q_{in}}{L}$

$位L=Q_{in}$ (show $位L$)

${\mathrm{桅}}_{net}=\frac{Q_{in}}{{系}_0}$ (put $位L$)