New Notes Blog: Find the potential a distance `z` above the center of a uniform line of charge with charge density `lambda ` and length `2L` .

Friday, 13 June 2014

Find the potential a distance $\displaystyle{z}$ above the center of a uniform line of charge with charge density $\displaystyle\lambda$ and length $\displaystyle{2}{L}$ .

For a continuous potential to equation is

$\displaystyle{V}=\int\frac{{{k}\cdot{d}{q}}}{{r}}$

Where $\displaystyle{k}=\frac{{1}}{{{4}\pi\epsilon_{{0}}}}$ , $\displaystyle{d}{q}$ is an infinitesimal piece of charge, and $\displaystyle{r}$ is the distance away $\displaystyle{d}{q}$ is from the point we are measuring.

We know the charge per unit length $\displaystyle\lambda=\frac{{{d}{q}}}{{{d}{l}}}$ where $\displaystyle{d}{l}$ is an infinitesimal chuck of the wire. If we let the wire lay along the x-axis then $\displaystyle{d}{l}={\left.{d}{x}\right.}$ . Therefore,

$\displaystyle{d}{q}=\lambda{\left.{d}{x}\right.}$

Now, looking at the picture we have a...

For a continuous potential to equation is

$\displaystyle{V}=\int\frac{{{k}\cdot{d}{q}}}{{r}}$

$\displaystyle{d}{q}=\lambda{\left.{d}{x}\right.}$

Now, looking at the picture we have a right triangle. So we know $\displaystyle{r}=\sqrt{{{{x}}^{{2}}+{{z}}^{{2}}}}$ . Then the integral becomes,

$\displaystyle{V}={\int_{{-{L}}}^{{L}}}\frac{{{k}\cdot\lambda{\left.{d}{x}\right.}}}{\sqrt{{{{x}}^{{2}}+{{z}}^{{2}}}}}={k}\cdot\lambda{\int_{{-{L}}}^{{L}}}\frac{{\left.{d}{x}}}{\sqrt{{{{x}}^{{2}}+{{z}}^{{2}}}}}$

From a table of integrals we can find that

$\displaystyle{V}={k}\lambda{\ln{{\left({x}+\sqrt{{{{x}}^{{2}}+{{z}}^{{2}}}}\right)}}}{{\mid}_{{-{L}}}^{{L}}}$

$\displaystyle{V}=\frac{\lambda}{{{4}\pi\epsilon_{{0}}}}{\ln{{\left(\frac{{{L}+\sqrt{{{{L}}^{{2}}+{{z}}^{{2}}}}}}{{-{L}+\sqrt{{{{L}}^{{2}}+{{z}}^{{2}}}}}}\right)}}}$

To simplify multiply the numerator and denominator by $\displaystyle{L}+\sqrt{{{{L}}^{{2}}+{{z}}^{{2}}}}$

$\displaystyle{V}=\frac{\lambda}{{{4}\pi\epsilon_{{0}}}}{\ln{{\left[\frac{{{\left({L}+\sqrt{{{{L}}^{{2}}+{{z}}^{{2}}}}\right)}}^{{2}}}{{-{{L}}^{{2}}+{{L}}^{{2}}+{{z}}^{{2}}}}\right]}}}$

$\displaystyle{V}={2}\cdot\frac{\lambda}{{{4}\pi\epsilon_{{0}}}}{\ln{{\left[\frac{{{L}+\sqrt{{{{L}}^{{2}}+{{z}}^{{2}}}}}}{{z}}\right]}}}$

$\displaystyle{V}=\frac{\lambda}{{{2}\pi\epsilon_{{0}}}}{\ln{{\left[\frac{{{L}+\sqrt{{{{L}}^{{2}}+{{z}}^{{2}}}}}}{{z}}\right]}}}$

This is the answer.

New Notes Blog

Friday, 13 June 2014

Find the potential a distance $\displaystyle{z}$ above the center of a uniform line of charge with charge density $\displaystyle\lambda$ and length $\displaystyle{2}{L}$ .

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Friday, 13 June 2014

Find the potential a distance above the center of a uniform line of charge with charge density and length .

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

Find the potential a distance $\displaystyle{z}$ above the center of a uniform line of charge with charge density $\displaystyle\lambda$ and length $\displaystyle{2}{L}$ .

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?