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phy142:lectures:5 [2012/02/04 16:33] mdawber [Lecture 5 - Electric Potential] |
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~~SLIDESHOW~~ | ~~SLIDESHOW~~ | ||

- | ====== Lecture 5 - Electric Potential ====== | + | ====== Lecture 5 - Calculating Electric Fields ====== |

+ | ---- | ||

+ | If you need a pdf version of these notes you can get it [[http://www.ic.sunysb.edu/class/phy141md/lecturepdfs/142lecture5S13.pdf|here]] | ||

+ | ===== Video of Lecture ===== | ||

- | ---- | ||

- | If you need a pdf version of these notes you can get it [[http://www.ic.sunysb.edu/class/phy141md/lecturepdfs/142lecture5S12.pdf|here]] | ||

- | ===== Video of lecture ==== | ||

- | |||

- | The video should play in any browser, but works best in anything that isn't Internet Explorer. If you are having trouble watching the video within the page you can [[http://www.ic.sunysb.edu/class/phy141md/lecturevids/phy142lecture5S2012.mp4|download the video]] and play it in [[http://www.apple.com/quicktime/download/|Quicktime]]. | ||

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- | ===== Electrical potential energy ===== | + | ===== Electric field from charge ===== |

- | In PHY 141 when we discussed [[phy141:lectures:12|conservation of energy]] we defined potential energy for a conservative force to be | ||

+ | The field from a point charge is a vector $\vec{E}$ which has magnitude | ||

- | $\Delta U = -\int_{1}^{2}\vec{F}\cdot\,d\vec{l} = - W_{Net} $ | + | $E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}$ |

- | In the case where a charge is moved a distance $d$ by a uniform electric field $E$ the work done is | + | and when $Q$ is positive points radially outwards from the charge. |

- | $W=Fd=qEd$ | + | You could write this as |

- | we can note here that this became a simple product because the charge moved in the direction of the field (as we can expect it should do if the only force acting is the electric force) | + | $\vec{E}=E\hat{r}$ |

- | and the change in the electric potential energy is | + | where $\hat{r}$ is a unit vector pointing away from the charge. |

- | $\Delta U=-W=-qEd$ | + | For a negative charge, Q is negative so the multiplication of the negative scalar with the outward $\hat{r}$ vector results in a field $\vec{E}$ that points inward towards the charge. |

- | ===== Electric potential ===== | + | ===== Field from an element of charge ===== |

- | In the same way it was useful to define electric field as the electric force divided by the charge it is also useful to define electric potential. | ||

- | If we define the electric potential energy at a point as $U$ then the electric potential at that point $V$ is | + | The field from an element of charge, |

- | $V=\frac{U}{q}$ | + | $\vec{dE}=dE\,\hat{r}$ |

- | In the example we just considered where a charge was moved a distance $d$ by a field $E$ | + | which also points radially outwards from the charge element has magnitude |

- | $\Delta U=-W=-qEd->\Delta V=-Ed$ | + | $dE=\frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^{2}}$ |

- | We should note that we have been careful to consider differences in potential, as with other forms of potential energy, the zero of electric potential energy, and hence electric potential, can be set arbitrarily. | ||

- | The unit for electric potential is the volt $V$ and we can see that as the volt is derived from dividing energy by charge $1V=1\frac{J}{C}$ | + | If we want to find the field directly from a distribution of charge we can integrate over the charged distribution, but we have to be careful to do take in to account the vector nature of the field! In principle if we want to find the total field vector we need to integrate taking into account the direction. In a 3 dimensional coordinate system with no symmetry this would be 3 integrals, one for each vector in our coordinate system and each vector would contain some relationship to define the component of field that points from the charge we are looking at to the point where we want to know the field. |

- | We can also see from the equation above that an equivalent unit for field to $\mathrm{N/C}$ is $\mathrm{V/m}$. | + | We will **not** do general problems like this because they are way too hard. |

- | ===== Signs ===== | + | We will only tackle problems using direct integration of charge to obtain field where there is cancellation of components of the field from different parts of the charge distribution. |

- | We should note that potential is defined with positive charges in mind, ie. a positive charge at a high potential with respect to some other point has a high potential energy with respect to that point, whereas a negative charge at a high potential with respect to some other point has a low potential energy with respect to that point. | + | ===== What needs to go inside the integral? ===== |

- | This information is contained in the equation | ||

- | $\Delta U = q\Delta V$ | + | Confusion often arises over what is constant and what needs to be integrated over. When calculating fields by direct integration using the integral $E=\int dE=\frac{1}{4\pi\epsilon_{0}}\int\frac{dq}{r^{2}}$ the question we need to ask is, does the distance of the charge from the point where we want to know the field change as we move in the charge distribution? If so we need to express the distance $r$ from the charge in terms of the variable over which we are integrating (see the example for the [[phy142:lectures:2&#long_line_of_charge|line of charge]]). If not, $\frac{1}{r^{2}}$ can be moved outside the integral. |

- | ===== Producing electric potential ===== | + | This was the key difference between the [[http://phy141md.local/doku.php?do=export_s5&id=phy142:lectures:3#slide2|ring of charge integral]] and the [[http://phy141md.local/doku.php?do=export_s5&id=phy142:lectures:3#slide3|line of charge integral]]. |

- | In a [[wp>Battery_(electricity)|battery]] chemical potential energy is used to create a potential difference between two terminals. As the battery runs down the potential difference stays fairly constant, but the amount of charge available decreases. There is a fixed amount of chemical potential energy available to be converted to electrical potential energy. | ||

- | A [[wp>Wimshurst_machine|Wimshurst machine]] allows the conversion of mechanical potential energy into a large electrical potential. The amount of charge separated is not large, so while there is a very large potential difference between the two terminals the amount of electrical potential energy produced by the machine is not very large. To understand why the potential is so large we will need to discuss capacitance (next week!). You can find a nice explanation of how the machine works [[http://www.coe.ufrj.br/~acmq/whyhow.html|here]] | + | ===== Integration over a charge distrbution ===== |

- | The spark occurs when the electric field between the two terminals exceeds the field for [[wp>Electrical_breakdown|electrical breakdown]] of the air between them. The breakdown field for air is approximately $3\times10^{6}\mathrm{V/m}$. | ||

- | ===== Electric potential and field ===== | + | Suppose we want to integrate over a distribution of charge. What does $\int dq$ really mean? |

- | The electric potential and field are easily related. | + | For a line of charge $\int dq = \int \lambda\,dx$ \\ |

+ | For a sheet of charge $\int dq = \int \sigma\,dA$ \\ | ||

+ | For volume of charge $\int dq =\int \rho\,dV$ | ||

- | $U_{b}-U_{a} = -\int_{a}^{b}\vec{F}\cdot\,d\vec{l} $ | + | If the charge density is uniform with respect to the variable being integrated over, then it can be taken out of the integral, if not then it needs to be included in the integral. |

- | $V_{b}-V_{a}=\frac{U_{b}-U_{a}}{q}=-\int_{a}^{b}\vec{E}\cdot\,d\vec{l} $ | + | Depending on the symmetry of the problem we might choose to integrate over Cartesian coordinates or polar coordinates, so for example, |

- | In the special case of a uniform electric field where the distance between points $a$ and $b$ is $d$ | + | for a rectangular sheet of charge $\int dq = \int \sigma\,dA=\int \int \sigma \,dx\,dy$\\ |

+ | for a circular sheet of charge $\int dq = \int \sigma\,dA=\int \int \sigma r\,dr\,d\theta$\\ | ||

- | $V_{b}-V_{a}=-\int_{a}^{b}\vec{E}\cdot\,d\vec{l}=-E\int_{a}^{b}dl=-Ed $ | + | Typically we have considered situations where one of these integrals is already done, for example when considering a uniform disk we first did the integral over $\theta$ for a given radius ([[http://phy141md.local/doku.php?do=export_s5&id=phy142:lectures:3#slide2|integral over a ring]]) and then integrated this result over the radius for a [[http://phy141md.local/doku.php?do=export_s5&id=phy142:lectures:3#slide4|disk]]. |

- | When the field is not uniform we can obtain the potential from the field by integration. | + | ===== Gauss's Law ===== |

+ | | ||

+ | $\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$ | ||

+ | Gauss's law will often give us a better way to evaluate the field for a distribution of charge, because on the right hand side we only need to evaluate the total amount of charge, we don't need to consider the direction or the distance of the field from the charge to the point at which we need to know the field. | ||

- | ===== Potential due to a point charge ===== | + | The most important step is to choose an appropriate Gaussian surface which reflects the symmetry of the charge. We should remember that the Gaussian surface must be a closed surface, and so it may be composed of more than one surface. If possible we want to choose the surfaces so that the field on each surface is uniform and we don't have to actually any integration over the surface. |

+ | It's important to remember that flux has a sign - it is positive when it is leaving the volume, and negative when entering (as the $\vec{A}$ always points out). | ||

- | If we want to find the change in potential in going from a distance $r_{a}$ to a distance $r_{b}$ from a single point charge we need to evaluate the integral | + | In this course we will consider only 3 kinds of symmetry when applying Gauss's Law. |

- | $V_{b}-V_{a}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}$ | ||

- | We can recall that the field is | + | ===== Spherical symmetry ===== |

+ | | ||

+ | | ||

+ | {{sphericalgausssurface.png}} | ||

+ | | ||

+ | If the field points out radially from the center of the sphere | ||

+ | | ||

+ | $\oint\vec{E}.d\vec{A}=\oint E\,dA$ | ||

+ | | ||

+ | and if it is uniform over the surface | ||

+ | | ||

+ | | ||

+ | $\oint E\,dA=E \oint dA=E4\pi r^{2}$ | ||

+ | | ||

+ | {{sphericalshell.png}} | ||

+ | | ||

+ | To find the enclosed charge | ||

+ | | ||

+ | $Q=\int\int\int\rho\,dV$ | ||

+ | | ||

+ | In cases where $\rho$ is either constant or only depends on r | ||

+ | | ||

+ | For a spherical volume the integral $\int\int\int\rho\,dV=\int\rho A dr=\int\rho 4\pi r^{2}dr$ | ||

+ | | ||

+ | ===== A spherical example ===== | ||

+ | | ||

+ | {{phy142m1fig2.png?300}} | ||

+ | | ||

+ | A insulating sphere of radius $r_{1}$ has a total positive charge Q which is evenly distributed throughout, except for in the center of the sphere where there is a spherical cavity of radius $r_{0}$ which contains no charge. Let $r$ be the distance of a point from the center of the sphere. Write your answers in terms of $Q$, $r$, $r_{1}$, $r_{0}$ and $k$ or $\varepsilon_{0}$. | ||

+ | | ||

+ | * What is the electric field $E(r)$ for points $r<r_{0}$? | ||

+ | | ||

+ | $E=0$ | ||

+ | | ||

+ | * What is the electric field $E(r)$ for points $r_{0}<r<r_{1}$? | ||

+ | | ||

+ | $E4\pi r^2=\frac{1}{\varepsilon_{0}}\int_{r_{0}}^{r}\rho 4\pi r^{2}\,dr=\frac{\rho}{\varepsilon_{0}}\frac{4\pi}{3}(r^{3}-r_{0}^3)$ | ||

+ | | ||

+ | $E=\frac{\rho}{\varepsilon_{0}}\frac{1}{3r^{2}}(r^{3}-r_{0}^3)$ | ||

+ | | ||

+ | $\rho=\frac{Q}{\frac{4}{3}\pi r_{1}^{3}-\frac{4}{3}\pi r_{0}^{3}}$ | ||

+ | | ||

+ | $E=\frac{1}{4\pi\varepsilon_{0}r^{2}}Q\frac{r^{3}-r_{0}^{3}}{r_{1}^3-r_{0}^{3}}=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r_{1}^{3}-r_{0}^{3}}(r-\frac{r_{0}^{3}}{r^{2}})$ | ||

+ | | ||

+ | * What is the electric field $E(r)$ for points $r>r_{1}$? | ||

$E=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}}$ | $E=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}}$ | ||

- | If we integrate along a path which is directly radially outward $\vec{E}$ and $d\vec{l}$ are in the same direction and | + | * Make a rough sketch of the electric field $E(r)$ against $r$. Make sure to label $r_{0}$ and $r_{1}$ on your sketch. |

- | $-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}=-\frac{Q}{4\pi\varepsilon_{0}}\int_{r_{a}}^{r_{b}}\frac{1}{r^{2}}dr=\frac{Q}{4\pi\varepsilon_{0}}[\frac{1}{r}]_{r_{a}}^{r_{b}}=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}})$ | + | {{spherewithcavityfieldr.png}} |

- | It is common to define the potential as being zero at $\infty$, and if we do this here (setting $r_{a}=\infty$) the potential as a function of the distance $r$ from a point charge is | + | ===== Cylindrical symmetry ===== |

- | $V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}$ | ||

- | ===== Work done in bringing two charges together ===== | ||

- | We just showed that the potential due to a point charge as a function of the distance from the charge is | + | {{linechargegauss.png}} |

- | $V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}$ | ||

- | If I have another charge $q$ that I want to move from a distance $r_{a}$ to a distance $r_{b}$ then the change in the potential is | + | If we consider a field that has cylindrical symmetry so that the flux through the end circular surfaces is zero and for the curved surface |

- | $V_{b}-V_{a}=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}})$ | + | $\oint\vec{E}.d\vec{A}=\oint E\,dA$ |

- | and the work done is | + | As the field is uniform over the surface |

- | $W=q(V_{b}-V_{a})=q(\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}}))$ | + | $\oint E\,dA=E \oint dA=E2\pi rl$ |

- | We should note that here we have taken into account the fact that the field through which are moving the charge is not uniform, we cannot simply say that $W=qEd$ | + | {{cylindricalshells.png}} |

- | ===== Charged conducting sphere ===== | + | To find the enclosed charge |

- | If we consider a conducting sphere of radius $r_{0}$ that carries a total charge of | + | $Q=\int\int\int\rho\,dV$ |

- | $Q$ on it's surface we can easily [[phy142:lectures:4&#charged_metal_sphere|determine from Gauss's law]] that | + | |

- | $E=0$<html>                </html> $[r<r_{0}]$\\ | + | In cases where $\rho$ is either constant or only depends on r |

- | $E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2}$<html>          </html> $[r\geq r_{0}]$ | + | |

- | As we did in the case of a point charge the integral $V_{b}-V_{a}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}$ can be done along a radial path so that $\vec{E}$ and $d\vec{l}$ are in the same direction. | + | $Q=\int\int\int\rho\,dV=\int\rho A dr=\int\rho 2\pi r l dr$ |

- | If we restrict our integral to points outside the sphere | ||

- | $-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}=-\frac{Q}{4\pi\varepsilon_{0}}\int_{r_{a}}^{r_{b}}\frac{1}{r^{2}}dr=\frac{Q}{4\pi\varepsilon_{0}}[\frac{1}{r}]_{r_{a}}^{r_{b}}=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}})$ | + | ===== A cylindrical example ===== |

- | and as before define the potential as being zero at $\infty$ | + | {{phy142finalfig1.png}} |

- | $V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}$<html>          </html> $[r>r_{0}]$ | + | A long cylinder of charge has a uniform volume charge density $\rho$ and radius $R_{1}$. Give your answers to the following questions in terms of some or all of the variables $\rho$, $r$, $ \varepsilon_{0}$, $R_{1}$, and $R_{2}$. |

- | This implies that at $r=r_{0}$ | + | A. Write an expression for the electric field as a function of the distance from the center of the cylinder, $r$, for $0<r \leq R_{1}$. |

- | $V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r_{0}}$ | + | $E2\pi rl=\int_{0}^{r}\rho 2\pi r l dr=\frac{\rho \pi r^{2} l}{\varepsilon_{0}}$ |

- | and as inside the sphere $E=0$ the integral $-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}=0$ which implies that everywhere inside the sphere | + | $E=\frac{\rho r}{2\varepsilon_{0}}$ |

- | $V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r_{0}}$<html>          </html> $[r\leq r_{0}]$ | + | B. Write an expression for the electric field as a function of the distance from the center of the cylinder, $r$, for $r>R_{1}$. |

- | Indeed, we can say generally that in addition to the field inside a conductor being everywhere zero, the potential at every point in conductor is everywhere the same. | + | $E2\pi rl=\int_{0}^{R_{1}}\rho 2\pi r l dr=\frac{\rho \pi R_{1}^{2} l}{\varepsilon_{0}}$ |

- | ===== Potential directly from charge ===== | + | $E=\frac{\rho R_{1}^{2}}{2r\varepsilon_{0}}$ |

- | + | ||

- | Whenever we have determined the electric field as a function of position we should be able to calculate the electric potential. Sometimes however the vector nature of the field can make this tricky. One of the great appeals of electric potential is that it is a scalar quantity, and therefore it can be easier to find the potential due to multiple charges or a distribution of charges than it is to find the field. | + | C. Plot $E$ as a function of $r$ for the situation considered in (a) and (b). |

- | {{potentialfrom2charges.png}} | + | There is no change from B. |

- | For example if I want to calculate the field due to equal positive and negative charges separated by some distance I need to pay attention to directions of the field, whereas to find the potential I simply add together the potential of each charge. We denote the negative charge as $A$ and the positive charge as $B$ and we can then say that the potential is given by | + | $E=\frac{\rho R_{1}^{2}}{2r\varepsilon_{0}}$ |

- | $V=\frac{1}{4\pi\varepsilon_{0}}(-\frac{Q}{R_{A}}+\frac{Q}{R_{B}})$ | ||

- | We can note that anywhere on the plane that lies in between the two charges the potential will be equal to zero. As each point on this plane has the same potential we call it an equipotential surface. Although the equipotential surface we have just considered is the most trivial to identify we could in fact identify an infinite number of equipotential surfaces (however, none of the others will be flat!). | + | D. I wish to surround the cylinder by a thin cylindrical sheet of charge such that there is no electric field outside the sheet. The sheet has a radius $R_{2}>R_{1}$ and has an area charge density $\sigma$ . Write an expression for the required area charge density $\sigma$ so that the field for $R>R_{2}$ is zero. |

- | We can take a closer look at the equipotetials in [[http://www.falstad.com/vector3de/|Falstad's 3D applet]]. (Note: it looks like this applet cannot plot the surfaces in 3D, but we can look at slices). | + | For there to be no enclosed charge |

- | ===== Potential due to a continuous distribution of charge ===== | + | $\rho\pi R_{1}^{2}l+\sigma 2\pi R_{2}l=0$ |

- | From the previous example we can conclude that for $n$ point charges each which produces a potential $V_{i}$ at point A the potential at point $A$ will be | + | $\sigma=-\frac{\rho R_{1}^{2}}{2R_{2}}$ |

- | $V_{A}=\sum\limits_{i=1}^{n}V_{i}=\frac{1}{4\pi\varepsilon_{0}}\sum\limits_{i=1}^{n}\frac{Q_{i}}{r_{ia}}$ | ||

- | Now if we want to instead consider a continuous charge distribution | + | E. For the case considered in D., write an expression for the electric field as a function of the distance from the center of the cylinder, $r$, for $R_{1}<r<R_{2}$. |

+ | ===== Planar symmetry ===== | ||

- | $V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}$ | ||

- | ===== Potential due to a ring of charge ===== | + | {{planeofchargegauss.png}} |

- | We can now consider a thin ring of charge, total charge $Q$ with radius $R$. We can ask what the potential is at a point on the axis going through the center of the ring a distance $x$ from the center of the ring. This situation is quite easy because all the charge is at the same distance from the point. So that the integral | + | If the plane is infinite then the flux through the curved surfaces is zero and we only need to be concerned about the flux through the flat prism surfaces of area $A$. The flux through those two faces will not always be the same, so in general if we define a field through the left face $\vec{E}_{L}$ and the right face $\vec{E}_{R}$ |

- | $V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}=\frac{1}{4\pi\varepsilon_{0}}\frac{1}{(x^{2}+R^{2})^{1/2}}\int dq=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{(x^{2}+R^{2})^{1/2}}$ | ||

- | {{ringofchargepotential.png}} | + | $\Phi_{E}=\oint\vec{E}\cdot d\vec{A}=\vec{E}_{L}\cdot\vec{A}+\vec{E}_{R}\cdot\vec{A}$ |

- | ===== On Monday ===== | + | {{cylindricalslice.png}} |

- | * More examples on how to calculate potential | + | |

- | * How to get field from potential | + | |

- | * A closer look at electrostatic potential energy | + | |

+ | $Q=\int\int\int\rho\,dV=\int\rho(l) A dl$ | ||

+ | or when there is a sheet of charge $\sigma$ | ||