Differences

This shows you the differences between the selected revision and the current version of the page.

whitepapers:sequences:emsequence:start 2019/07/22 11:52 whitepapers:sequences:emsequence:start 2019/07/22 11:56 current
Line 8: Line 8:
A more in-depth discussion of the rationale, student thinking, and the way these activities fit together can be found in a discussion of how these activities break the learning into manageable [[.:pieces]] A more in-depth discussion of the rationale, student thinking, and the way these activities fit together can be found in a discussion of how these activities break the learning into manageable [[.:pieces]]
 +
 +
Line 14: Line 16:
Students may be familiar with the iconic equation for the electric potential (due to a point charge): Students may be familiar with the iconic equation for the electric potential (due to a point charge):
$$\text{Iconic:} \qquad V=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$$ $$\text{Iconic:} \qquad V=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$$
-With information about the type of source distribution, one can write or select the appropriate coordinate independent equation for $V$. For example, if the source is a line of charge:+With information about the type of source distribution, one can write or select the appropriate coordinate independent equation for $V$. For example, if the source is a linear (i.e. 1 dimensional):
$$\text{Coordinate Independent:} \qquad V=\frac{1}{4 \pi \epsilon_0} \int\frac{\lambda | d\vec r' |}{| \vec r - \vec r' |}$$ $$\text{Coordinate Independent:} \qquad V=\frac{1}{4 \pi \epsilon_0} \int\frac{\lambda | d\vec r' |}{| \vec r - \vec r' |}$$
Looking at symmetries of the source, one can choose a coordinate system and write the equation for the potential in terms of this coordinate system. Note that this step is often combined with the following step, though one may wish to keep them separate for the sake of careful instruction. Looking at symmetries of the source, one can choose a coordinate system and write the equation for the potential in terms of this coordinate system. Note that this step is often combined with the following step, though one may wish to keep them separate for the sake of careful instruction.
$$\text{Coordinate Dependent:} \qquad V=\frac{1}{4 \pi \epsilon_0} \int\frac{\lambda |ds'\ \hat s + s'\ d\phi'\ \hat \phi + dz'\ \hat z|}{| s'^2 + s^2 +2ss' \cos(\phi-\phi') + z^2|}$$ $$\text{Coordinate Dependent:} \qquad V=\frac{1}{4 \pi \epsilon_0} \int\frac{\lambda |ds'\ \hat s + s'\ d\phi'\ \hat \phi + dz'\ \hat z|}{| s'^2 + s^2 +2ss' \cos(\phi-\phi') + z^2|}$$
Using what you one about the geometry of the source, one can simplify the expression. For example, if the source is a ring of charge with radius $R$ and charge $Q$ in the $x$-, $y$-plane the integral becomes: Using what you one about the geometry of the source, one can simplify the expression. For example, if the source is a ring of charge with radius $R$ and charge $Q$ in the $x$-, $y$-plane the integral becomes:
-$$\text{Coordinate and Geometry Dependent:} \qquad V=\frac{1}{4 \pi \epsilon_0} \int\frac{\lambda s'\ d\phi'}{| s'^2 + R^2 +2Rs' \cos(\phi-\phi') + z^2|}$$+$$\text{Coordinate and Geometry Dependent:} \qquad V=\frac{1}{4 \pi \epsilon_0}\, \frac{Q}{2\pi R} \int_0^{2\pi}\frac{R\ d\phi'}{| R^2 + s^2 +2Rs \cos(\phi-\phi') + z^2|}$$
====Lecture: Chop, Calculate, and Add==== ====Lecture: Chop, Calculate, and Add====

Personal Tools