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courses:order20:vforder20:vfcalculating 2019/04/11 12:40 | courses:order20:vforder20:vfcalculating 2019/05/30 08:08 current | ||
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===== In-class Content ===== | ===== In-class Content ===== | ||

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+ | ====Lecture: Electric Potential==== | ||

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+ | Students may be familiar with the iconic equation for the electric potential (due to a point charge): | ||

+ | $$\text{Iconic:} \qquad V=\frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$$ | ||

+ | 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: | ||

+ | $$\text{Coordinate Independent:} \qquad V=\frac{1}{4 \pi \epsilon_0} \int\frac{\lambda | d\vec r' |}{| \vec r - \vec r' |}$$ | ||

+ | 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. | ||

+ | $$\text{Coordinate Dependent:} \qquad V=\frac{1}{4 \pi \epsilon_0} \int\frac{\lambda |ds'\ \hat s + s'\ d\phi'\ \hat \phi + dz'\ \hat z|}{| s'^2 + s^2 +2ss' \cos(\phi-\phi') + z^2|}$$ | ||

+ | Using what you one about the geometry of the situation, one can possibly simplify the numerator. For example: | ||

+ | $$\text{Coordinate and Geometry Dependent:} \qquad V=\frac{1}{4 \pi \epsilon_0} \int\frac{\lambda s'\ d\phi'}{| s'^2 + s^2 +2ss' \cos(\phi-\phi') + z^2|}$$ | ||

+ | Emphasize that "primes" (i.e., $s'$, $\phi'$, $z'$, etc.) are used to indicate the location of charge in the charge distribution. | ||

====Lecture: Chop, Calculate, and Add==== | ====Lecture: Chop, Calculate, and Add==== | ||

* To find the area under a curve, one may chop up the x-axis into small pieces (of width $dx$). The area under the curve is then found by calculating the area for each region of $dx$ (which is $f(x) dx$) and then summing up all of those areas. In the limit where $dx$ is small enough, the sum becomes an integral. | * To find the area under a curve, one may chop up the x-axis into small pieces (of width $dx$). The area under the curve is then found by calculating the area for each region of $dx$ (which is $f(x) dx$) and then summing up all of those areas. In the limit where $dx$ is small enough, the sum becomes an integral. | ||

- | * One could also find the area under a curve by chopping up both the x- and y-axes (chop), calculating the area of each small area under the curve (calculate), and adding all of those together with a double sum or double integral (add). | + | {{courses:order20:vforder20:chop_x.png?300|}} |

+ | * One could also find the area under a curve by chopping up both the x- and y-axes (chop), calculating the area of each small area under the curve (calculate), and adding all of those together with a double sum or double integral. | ||

+ | {{courses:order20:vforder20:chop_x_y.png?300|}} | ||

* This approach can be used to find the area of a cone, where the 'horizontal' length of each area is $r d\phi$ and the 'vertical' length is $dr$, giving an area of $dA = r d\phi dr$. It is important to make sure that the limits of integration are appropriate so that the integrals range over the whole area of interest. | * This approach can be used to find the area of a cone, where the 'horizontal' length of each area is $r d\phi$ and the 'vertical' length is $dr$, giving an area of $dA = r d\phi dr$. It is important to make sure that the limits of integration are appropriate so that the integrals range over the whole area of interest. | ||

- | * If one wants to calculate something other than length, area, or volume, such as if one sprinkled charge over a thin bar, then chop, calculate, and add still works. Again, chop the bar up into small lengths of $dx$. Then calculate the change $dQ$ on each length ($dQ = \lambda dx$), and add all of the $dQ$s together in a sum or integral. | + | {{courses:order20:vforder20:dA_for_cone.png?200|}} |

+ | * If one wants to calculate something other than length, area, or volume, such as if one sprinkled charge over a thin bar, then chop, calculate, and add still works. Again, chop the bar up into small lengths of $dx$. Then calculate the charge $dQ$ on each length ($dQ = \lambda dx$), and add all of the $dQ$s together in a sum or integral. | ||

+ | {{courses:order20:vforder20:chop_lambda.png?300|}} | ||

*This also works for calculating something (such as charge) over a volume. For a thick cylindrical shell with a charge density $\rho(\vec r)$, chop the shell into small volumes of $d \tau$ (which will be a product of 3 small lengths, e.g. $d \tau = r d\phi\ dr\ dz$), multiply this volume by the charge density at each part of the shell (defined by e.g. $r, \phi,$ and $z$), and add the resulting $dQ$s together. | *This also works for calculating something (such as charge) over a volume. For a thick cylindrical shell with a charge density $\rho(\vec r)$, chop the shell into small volumes of $d \tau$ (which will be a product of 3 small lengths, e.g. $d \tau = r d\phi\ dr\ dz$), multiply this volume by the charge density at each part of the shell (defined by e.g. $r, \phi,$ and $z$), and add the resulting $dQ$s together. | ||

+ | {{courses:order20:vforder20:chop_rho.png?200|}} | ||

====Activities==== | ====Activities==== | ||

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* [[..:..:activities:vfact:vfvring|Electrostatic potential due to a ring of charge]] (SGA - 50 min) | * [[..:..:activities:vfact:vfvring|Electrostatic potential due to a ring of charge]] (SGA - 50 min) | ||

- | * [[..:..:activities:vfact:vfvring|Series expansion of potential due to a ring of charge ]] (Extension of previous SGA + 20-30 min) | ||

* [[..:..:lecture:vflec:vflines|Lines of Charge]] (Lecture: 30 min) | * [[..:..:lecture:vflec:vflines|Lines of Charge]] (Lecture: 30 min) | ||