Notes: Calculate the electric field above a ring of charge of radius `R` and charge density `lambda` at a point `P` centered on the axis of the ring.

Sunday, 7 December 2014

Calculate the electric field above a ring of charge of radius $\displaystyle{R}$ and charge density $\displaystyle\lambda$ at a point $\displaystyle{P}$ centered on the axis of the ring.

In general, the contribution to the total electric field $\displaystyle{E}$ from an infinitesimal piece of charge $\displaystyle{d}{E}$ is

$\displaystyle{d}{E}=\frac{{{k}\cdot{d}{q}}}{{{r}}^{{2}}}$

The total field is then calculated by integrating over all the charge:

$\displaystyle{E}=\int\frac{{{k}\cdot{d}{q}}}{{{r}}^{{2}}}$

Where $\displaystyle{k}=\frac{{1}}{{{4}\pi\epsilon_{{0}}}}$ and $\displaystyle{r}$ is the distance from $\displaystyle{d}{q}$ to the point $\displaystyle{P}$ .

For this problem, if the point $\displaystyle{P}$ is a distance $\displaystyle{z}$ above the ring then we can set up a triangle to get a relationship between $\displaystyle{z}$ and...

In general, the contribution to the total electric field $\displaystyle{E}$ from an infinitesimal piece of charge $\displaystyle{d}{E}$ is

$\displaystyle{d}{E}=\frac{{{k}\cdot{d}{q}}}{{{r}}^{{2}}}$

The total field is then calculated by integrating over all the charge:

$\displaystyle{E}=\int\frac{{{k}\cdot{d}{q}}}{{{r}}^{{2}}}$

Where $\displaystyle{k}=\frac{{1}}{{{4}\pi\epsilon_{{0}}}}$ and $\displaystyle{r}$ is the distance from $\displaystyle{d}{q}$ to the point $\displaystyle{P}$ .

For this problem, if the point $\displaystyle{P}$ is a distance $\displaystyle{z}$ above the ring then we can set up a triangle to get a relationship between $\displaystyle{z}$ and the distance $\displaystyle{r}$ .

$\displaystyle{{r}}^{{2}}={{z}}^{{2}}+{{R}}^{{2}}$

The vector $\displaystyle{d}{E}$ pointing at point $\displaystyle{P}$ from a chunk of charge, can be broken up into two components, one in the radial direction and one in the vertical direction.

$\displaystyle{d}{E}={d}{E}_{{r}}+{d}{E}_{{z}}$

When considering the total electric field contribution from the ring, the radial contribution of every piece of charge will sum to zero and only the $\displaystyle{z}$ contribution will be left due to the symmetry of the ring.

$\displaystyle{d}{E}_{{z}}={d}{E}\cdot{\cos{{\left(\phi\right)}}}={d}{E}\cdot{\left(\frac{{z}}{{r}}\right)}=\frac{{{k}\cdot{d}{q}}}{{{r}}^{{2}}}\cdot{\left(\frac{{z}}{{r}}\right)}$

$\displaystyle{E}_{{z}}=\int\frac{{{k}\cdot{d}{q}}}{{{r}}^{{2}}}\cdot{\left(\frac{{z}}{{r}}\right)}$

$\displaystyle{E}_{{z}}=\int\frac{{{k}\cdot{z}\cdot{d}{q}}}{{{r}}^{{3}}}$

$\displaystyle\lambda$ is the piece of charge divided by a piece of length along the ring.

$\displaystyle\lambda=\frac{{{d}{q}}}{{{d}{l}}}$

From the fact that $\displaystyle{r}\cdot\theta$ is equal to the arc length around a circle we can say that

$\displaystyle{d}{l}={R}\cdot{d}{\left(\theta\right)}$

Therefore,

$\displaystyle{E}_{{z}}=\int\frac{{{k}\cdot{z}\cdot\lambda{d}{l}}}{{{r}}^{{3}}}$

$\displaystyle{E}_{{z}}=\int\frac{{{k}\cdot{z}\cdot\lambda{R}{d}{\left(\theta\right)}}}{{{r}}^{{3}}}$

Now substitute for $\displaystyle{r}$ and integrate theta from $\displaystyle{\left[{0},{2}\pi\right]}$ .

$\displaystyle{E}_{{z}}={\int_{{0}}^{{{2}\pi}}}\frac{{{k}\cdot{z}\cdot\lambda{R}{d}{\left(\theta\right)}}}{{{\left(\sqrt{{{{z}}^{{2}}+{{R}}^{{2}}}}\right)}}^{{3}}}$

$\displaystyle{E}_{{z}}=\frac{{{k}\cdot{z}\cdot\lambda{R}}}{{{\left(\sqrt{{{{z}}^{{2}}+{{R}}^{{2}}}}\right)}}^{{3}}}{\int_{{0}}^{{{2}\pi}}}{d}{\left(\theta\right)}$

$\displaystyle{E}_{{z}}=\frac{{{2}\pi{k}\cdot{z}\cdot\lambda\cdot{R}}}{{{\left(\sqrt{{{{z}}^{{2}}+{{R}}^{{2}}}}\right)}}^{{3}}}$

Therefore the electric field at point $\displaystyle{P}$ is entirely in the $\displaystyle{z}$ -direction with magnitude:

$\displaystyle{E}_{{z}}=\frac{{{z}\cdot\lambda\cdot{R}}}{{{2}\epsilon_{{0}}{{\left(\sqrt{{{{z}}^{{2}}+{{R}}^{{2}}}}\right)}}^{{3}}}}$

Notes

Sunday, 7 December 2014

Calculate the electric field above a ring of charge of radius $\displaystyle{R}$ and charge density $\displaystyle\lambda$ at a point $\displaystyle{P}$ centered on the axis of the ring.

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Sunday, 7 December 2014

Calculate the electric field above a ring of charge of radius and charge density at a point centered on the axis of the ring.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

Calculate the electric field above a ring of charge of radius $\displaystyle{R}$ and charge density $\displaystyle\lambda$ at a point $\displaystyle{P}$ centered on the axis of the ring.

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?