http://sites.science.oregonstate.edu/BridgeBook/ 2020-01-26T16:07:05-08:00 http://sites.science.oregonstate.edu/BridgeBook/ http://sites.science.oregonstate.edu/BridgeBook/lib/images/favicon.ico text/html 2019-02-07T15:20:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/320review?rev=1549581600 Here is a short list of some of the main topics covered so far: \begin{itemize}\item Power series (good approximation at a point) \item Curvilinear coordinates ($d\rr$; line, surface, and volume integrals) \item Finding the electrostatic potential $V(\rr)$ (from either $\rho$ or $\EE$) \item Finding the electric field $\EE(\rr)$ (from either $\rho$ or $V$) \end{itemize} text/html 2012-12-31T21:08:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/422review?rev=1357016880 \begin{itemize}\item Gauss's Law and Ampère's Law \item Divergence and Curl \item Divergence Theorem and Stokes' Theorem \item Curl-free and divergence-free vector fields \item Integrals to find $\AA$, the magnetic vector potential, and $\BB$, the magnetic field \item Integral (Gauss, Ampère) and Differential (Gauss, Ampère) form of Maxwell's Equations \item Boundary conditions on the electric and magnetic fields \item Relationship between potentials, fields, and sources (Electric, Ma… text/html 2019-02-07T15:10:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ampere?rev=1549581000 Amp\'ere's Law says that a geometric property of the magnetic field (namely the circulation around a particular closed loop) is the same as the current enclosed by that same loop (times $\mu_0$ in the system of units that we are using): \begin{eqnarray*} \oint\limits_{\rm loop} \BB \cdot d\rr = \mu_0 \, I_{\rm enclosed} \end{eqnarray*} text/html 2012-12-31T21:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ampereact?rev=1357017120 A steady current is flowing parallel to the axis through an infinitely long cylindrical shell of inner radius $a$ and outer radius $b$. Choose one or more of the current densities given below: (In each case, $\alpha$ and $k$ are constants with appropriate units.) text/html 2019-02-07T15:36:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/amperehint?rev=1549582560 text/html 2012-11-01T08:13:44-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/aring?rev=1351782824 Recall that for a thin current-carrying loop we have \begin{eqnarray*} \AA(\rr) = {\mu_0\over 4\pi} \int {\II(\rrp)\,ds'\over|\rr-\rrp|} \end{eqnarray*} Find an expression for the magnetic vector potential for a spinning circular ring with radius $R$, charge $Q$, and period $T$. The expression should be valid everywhere in space and simplified enough that it could be evaluated by a computer algebra programs such as Mathematica or Maple. text/html 2019-02-07T15:20:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/aringhint?rev=1549581600 Recall that for a thin current-carrying loop we have \begin{eqnarray*} \AA(\rr) = {\mu_0\over 4\pi} \int {\II(\rrp)\,ds'\over|\rr-\rrp|} = {\mu_0\over 4\pi} \int {\lambda(\rrp)\,\vv\,ds'\over|\rr-\rrp|} \end{eqnarray*} For a circular ring of current, we have \begin{eqnarray*} \vv = \frac{2\pi R}{T}\,\phat = \frac{2\pi R}{T}\, \Bigl( -\sin\phi\,\ii + \cos\phi\,\jj \Bigr) \end{eqnarray*} and \begin{eqnarray*} |\rr-\rrp|= \sqrt{r^2 - 2\, rR \cos(\phi-\phi') + R^2 + z^2} \end{eqnarray*} text/html 2012-11-01T07:55:17-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/bbound?rev=1351781717 Given a current-carrying surface, it makes sense to ask what the component $B_\perp$ of the magnetic field is perpendicular to the surface, which is $B_\perp = \BB\cdot\nn$, where $\nn$ is the unit normal to the surface. The component parallel to the surface, $B_{\parallel}$, is more subtle, since there are an infinite number of directions parallel to the surface. However, since for a current-carrying surface there is a preferred direction in the surface, namely the direction of the current $\… text/html 2012-11-02T10:04:30-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/bboundhint?rev=1351875870 In (ss)~{Activity: Boundary Conditions on Magnetic Fields}, you were asked to use an infinitesimally small Gaussian surface and an infinitesimally small Amperian loop to discover how the different components of the magnetic field $\BB$ behave as they cross a surface current. You should have chosen the surface and/or loop small enough that the field is effectively constant, except, of course, that there might be abrupt discontinuous changes where the current resides. text/html 2019-02-07T15:08:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/bplane?rev=1549580880 Consider the magnetic field due to a uniform current everywhere in the $xy$-plane, moving in the $x$-direction. In what direction is the magnetic field? The Biot-Savart Law says that \begin{eqnarray*} \BB = {\mu_0\over 4\pi} \int {\KK(\rrp)\times(\rr-\rrp)\,dA'\over|\rr-\rrp|^3} \end{eqnarray*} But $\KK\times(\rr-\rrp)$ is perpendicular to $\KK$, and since $\KK$ points (everywhere) in the $x$-direction, the $\xhat$-component of $\BB$ must vanish, that is, $B_x=0$. text/html 2012-11-02T11:10:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/bring?rev=1351879800 Recall that for a thin current-carrying loop we have \begin{eqnarray*} \BB(\rr) = {\mu_0\over 4\pi} \int {\II(\rrp)\times(\rr-\rrp)\,ds'\over|\rr-\rrp|^3} = - {\mu_0 I\over 4\pi} \int {(\rr-\rrp)\times d\rrp\over|\rr-\rrp|^3} \end{eqnarray*} Find an expression for the magnetic field for a spinning circular ring with radius $R$, charge $Q$, and period $T$. The expression should be valid everywhere in space and simplified enough that it could be evaluated by a computer algebra programs such as M… text/html 2019-02-07T15:20:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/bringhint?rev=1549581600 Recall that for a thin current-carrying loop we have \begin{eqnarray*} \BB(\rr) = {\mu_0\over 4\pi} \int {\II(\rrp)\times(\rr-\rrp)\,ds'\over|\rr-\rrp|^3} = - {\mu_0 I\over 4\pi} \int {(\rr-\rrp)\times d\rrp\over|\rr-\rrp|^3} \end{eqnarray*} For a circular ring of current, we have \begin{eqnarray*} |\rr-\rrp|= \sqrt{r^2 - 2\, rR \cos(\phi-\phi') + R^2 + z^2} \end{eqnarray*} text/html 2012-10-28T15:21:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/bs?rev=1351462860 We started with the superposition principle for the (electric) scalar potential \begin{eqnarray*} V(\rr) = {1\over 4\pi\epsilon_0} \int {\rho(\rrp)\,d\tau'\over|\rr-\rrp|} \end{eqnarray*} and the corresponding superposition principle for the electric field \begin{eqnarray*} \EE(\rr) = {1\over 4\pi\epsilon_0} \int {\rho(\rrp) (\rr-\rrp)\,d\tau'\over|\rr-\rrp|^3} \end{eqnarray*} But we also have the relation \begin{eqnarray*} \EE = - \grad V \end{eqnarray*} and we could have calculated the above … text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/compareab?rev=1277055518 * Steal results from HW. text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/concept?rev=1277055518 text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/conductors?rev=1277055518 Please read Section~2.5 in Griffiths. text/html 2015-08-19T10:45:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/coulomb?rev=1440006300 The electric field of a point charge at the origin is given by \begin{eqnarray*} \EE = \frac{1}{4\pi\epsilon_0} \frac{q\,\rhat}{r^2} \end{eqnarray*} We can take the divergence of this field using the expression in (ss)~{The Divergence in Curvilinear Coordinates} for the divergence of a radial vector field, which yields \begin{eqnarray*} \grad\cdot\EE = \frac{1}{r^2} \Partial{}{r}\Bigl(r^2 E_r\Bigr) = \frac{1}{4\pi\epsilon_0} \frac{q\,\rhat}{r^2} \frac{1}{r^2} \Partial{q}{r} = 0 \end{eqnarra… text/html 2015-08-18T21:55:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/cube?rev=1439960100 The integral in Gauss' Law does not depend on the shape of the surface being used. So let's replace the sphere in the example in (ss)~{Gauss's Law} with a cube. Suppose the charge is at the origin, and the length of each side of the cube is $2$. Start by computing the flux through one face. Using the relationship \begin{equation} \rhat=\frac{x\,\xhat+y\,\yhat+z\,\zhat}{r} \end{equation} we have \begin{eqnarray*} \hbox{flux} = \Int_{-1}^1\Int_{-1}^1 {1\over4\pi\epsilon_0} {q\,\rhat\over r^… text/html 2012-10-28T15:13:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/cubehint?rev=1351462380 (ss)~{Activity: Flux through a Cube} asks you to compute the flux of the electric field through the faces of a cube. If you use technology to examine the integrand, you will discover that the integrand is largest near the center of the face. Why? Yes, the center of the face is closest to the charge; is that the only reason? Don't forget that the direction of the electric field also affects the integral, not merely its magnitude. text/html 2012-12-31T11:46:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/curlact?rev=1356983160 The curl of a function of three variables is a vector at each point in space. How can we graph such vector fields? How many different ways can you represent this information? After you have thought about these questions yourself, you can use the Maple worksheet [Visualizing the Curl] or the Mathematica notebook [Visualizing the Curl] to explore several different mechanisms for visualizing the curl in two and three dimensions. text/html 2012-10-09T11:27:57-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/curlfree?rev=1349807277 A vector field $\FF$ is said to be curl free if any one of the following conditions holds: \begin{enumerate}\item $\grad\times\FF=\zero$; \item $\int\FF\cdot d\rr$ is independent of path; \item $\oint\limits\FF\cdot d\rr=0$ for any closed path; \item $\FF$ is the gradient of some scalar field, that is, $\FF=\grad f$ for some $f$. \end{enumerate} text/html 2019-02-07T10:36:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/curlfreehint?rev=1549564560 Claim: the following conditions are equivalent: \begin{enumerate}\item $\grad\times\FF=\zero$; \item $\int\FF\cdot d\rr$ is independent of path; \item $\oint\limits\FF\cdot d\rr=0$ for any closed path; \item $\FF$ is the gradient of some scalar field, that is, $\FF=\grad f$ for some $f$. \end{enumerate} text/html 2019-02-07T15:18:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/curlhint?rev=1549581480 Figure 1:Using round boxes to determine the curl. In (ss)~{Activity: Using Technology to Visualize the Curl} you were asked to apply the geometric definition of curl. It is important to realize that the divergence is a vector field; it is not enough to examine the behavior near a single point, such as the origin. text/html 2015-08-18T16:37:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/currents?rev=1439941020 How do we measure current? By counting the charges which go past. So use a stopwatch, and count. What are the units? Coulombs per second. What answer do we get? That depends on the (linear) charge density $\lambda$ and the velocity $\vv$. At any point in a wire, the current is given by \begin{eqnarray*} \II = \lambda\,\vv \end{eqnarray*} and the number of charges which go past per unit time is clearly just the magnitude, $I=|\II|$. Since the direction of the current is obvious in a wire,… text/html 2015-08-16T08:36:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/dimensions?rev=1439739360 During problem-solving in practical applications, it is a useful strategy to keep track of the units of the various quantities. When the desired solution is numerical, it is essential that we use a consistent system of units. In theoretical derivations, on the other hand, where the desired answer is an equation, we may not need to choose a particular system of units. Nevertheless, as a sense-making strategy, it can be valuable to keep track of some of the information contained in units, i.e.~t… text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/disk?rev=1277055518 What is the potential due to a uniformly charged disk? First of all, we need to generalize our previous expressions to surface densities. Chop up the region, determine the potential on each one, and add them up, obtaining \begin{eqnarray*} V(\rr) = \Int_{\hbox{disk}} text/html 2012-10-08T19:53:04-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/disk2?rev=1349751184 Recall that the electric field on a surface is given by \begin{eqnarray*} \EE(\rr) = \int \frac{1}{4\pi\epsilon_0} \frac{\sigma(\rrp)(\rr-\rrp)\,dA}{|\rr-\rrp|^3} \end{eqnarray*} Let's find the electric field due to a charged disk, on the axis of symmetry. text/html 2012-12-31T11:46:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/divact?rev=1356983160 The divergence of a function of three variables is a scalar at each point in space. You can compute the divergence using a formula, but you should also be able to predict the results of such computations directly from the geometric definition of divergence. text/html 2012-10-09T11:28:36-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/divfree?rev=1349807316 A vector field $\FF$ is said to be divergence free if any one of the following conditions holds: \begin{enumerate}\item $\grad\cdot\FF=0$; \item $\int\FF\cdot d\AA$ is independent of surface; \item $\oint\FF\cdot d\AA=0$ for any closed surface; \item $\FF$ is the curl of some other vector field, that is, $\FF=\grad\times\GG$ for some $\GG$. \end{enumerate} text/html 2015-08-27T18:02:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/divfreehint?rev=1440723720 Claim: the following conditions are equivalent: \begin{enumerate}\item $\grad\cdot\FF=0$; \item $\int\FF\cdot d\AA$ is independent of surface; \item $\oint\FF\cdot d\AA=0$ for any closed surface; \item $\FF$ is the curl of some other vector field, that is, $\FF=\grad\times\GG$ for some $\GG$. \end{enumerate} text/html 2019-02-07T15:18:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/divhint?rev=1549581480 Figure 1:Using round boxes to determine the divergence. In (ss)~{Activity: Using Technology to Visualize the Divergence}, you were asked to apply the geometric definition of divergence. It is important to realize that the divergence is a function; it is not enough to examine the behavior near a single point, such as the origin. text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/dotperp?rev=1277055518 A field can be expressed in many different coordinate systems. For example, in rectangular coordinates, the electric field is \begin{eqnarray*} \EE = E_x\,\ii + E_y\,\jj + E_z\,\kk \end{eqnarray*} Similarly, in cylindrical coordinates, \begin{eqnarray*} \EE = E_r\,\rr + E_\phi\,\phat + E_z\,\zhat \end{eqnarray*} What is the $x$-component of $\EE$? Surely just $E_x$. The $r$-component? $E_r$. text/html 2012-10-16T15:17:01-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/drawfieldact?rev=1350425821 Using only what you know about the relationship of charges to electrostatic electric fields, namely: $$\EE=\frac{1}{4\pi\epsilon_0} \, \frac{Q\hat{r}}{r^2}$$ and the superposition principle, sketch the electric field (the vector field) for each of the following static charge configurations: \begin{itemize}\item Four positive charges arranged in a square. \item Two positive charges and two negative charges arranged in a square, with like charges diagonally opposite each other. \item A line segme… text/html 2012-10-28T15:03:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/drawfieldhint?rev=1351461780 There are two main approaches to drawing the electric field vectors: \begin{itemize}\item Vector addition of the known electric fields due to each source; \item Computing the gradient of the potential, if known, for the given configuration. \end{itemize} text/html 2012-10-09T11:24:06-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/drawquadrupole?rev=1349807046 In introductory physics, you may have learned how to find electrostatic potentials due to several discrete sources by first finding the electric field. However, it is also possible, and often useful, to find electrostatic potentials directly. Using only what you know about the relationship of charges to electrostatic potentials, namely: $$V=\frac{1}{4\pi\epsilon_0} \, \frac{Q}{r}$$ and the superposition principle, sketch the equipotential surfaces for each of the following static charge config… text/html 2015-08-16T17:29:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/drawquadrupolehint?rev=1439771340 (ss)~{Activity: Visualization of Potentials} should have given you practice visualizing electrostatic potentials due to some simple charge configurations. Some things that you might have needed to pay attention to are: \begin{enumerate}\item The electrostatic potential is a scalar field, not a vector field, i.e. it is a number at every point in space, not a vector. \item The electrostatic potential at a given point due to several discrete charges is the scalar sum of the potentials due to the s… text/html 2012-10-30T15:07:58-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ebound?rev=1351634878 How does the electric field behave near a charged surface? There is no obvious reason for the electric field to be the same on both sides of the surface. Using an infinitesimally small Gaussian surface and an infinitesimally small Amperian loop, you should explore how the electric field changes across a surface charge density. Do not assume that the surface is flat or has any special symmetry. It will help to look separately at the component of the electric field parallel to the surface $\EE_… text/html 2012-11-02T10:07:41-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/eboundhint?rev=1351876061 In (ss)~{Activity: Boundary Conditions on Electric Fields}, you were asked to use an infinitesimally small Gaussian surface and an infinitesimally small Amperian loop to discover how the different components of the electric field $\EE$ behave as they cross a surface charge density. You should have chosen the surface and/or loop small enough that the field is effectively constant, except, of course, that there might be abrupt discontinuous changes where the charge resides. You should be able t… text/html 2015-08-17T16:53:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/electric?rev=1439855580 We define the electric and gravitational fields as \begin{eqnarray*} \gv &=& -\grad \Phi \\ \EE &=& -\grad V \end{eqnarray*} Masses and (positive) test charges will move in the direction of the corresponding field, that is, in the direction in which the potential decreases. text/html 2012-10-07T20:27:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/emenergyact?rev=1349666820 Imagine an empty room. One at a time, your friends enter this room, and take their seats. Pretend that each friend represents a charge, being brought in from infinity. As each charge is moved into its position, compute the energy (work) needed to bring this charge from its starting point to its final location. Write down an expression for the total energy when all charges have been moved into position. text/html 2012-10-07T21:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/emenergyhint?rev=1349669640 As you work through the computation of the total energy in the final charge configuration, you will quickly appreciate the need for a compact notation, in this case the summation notation. What is the potential when no charges are present? $V=0$, so no energy is needed to bring the first charge in. text/html 2012-10-29T12:31:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/energy?rev=1351539060 How much work is done in assembling a collection of $n$ point charges? Work is force times distance, which in this context takes the form \begin{eqnarray*} W = \int \FF\cdot d\rr = -q \int \EE\cdot d\rr = q \>\>\Delta V \end{eqnarray*} Moving the first charge requires no work --- since there is no electric field. The second charge needs to be moved in the (Coulomb) field of the first, the third in the field of the first two, and so on. Continuing in this manner, we see that the work done in as… text/html 2012-10-07T15:36:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ering1act?rev=1349649360 Find the electric field for a uniformly charged ring with total charge $Q$ and radius $R$. Then determine the power series expansions that represent the electric field due to the charged ring, both on axis and in the plane of the ring, and both near to and far from the ring. text/html 2019-02-07T15:17:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ering1hint?rev=1549581420 In (ss)~{Activity: Electric Field due to a Ring}, you will have found an integral expression for the electric field due to a uniform ring of charge, then used power series methods to approximate the integral in various regions. This activity has much in common with (ss)~{Activity: A Uniformly Charged Ring}, which you may want to review at this time. In addition to the ideas discussed in the wrap-up to that activity in (ss)~{Wrap-Up: A Uniformly Charged Ring}, you may have needed to pay atten… text/html 2019-02-07T15:19:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/etriangle?rev=1549581540 Starting with the electrostatic potential for a point charge, we used the superposition principle in~(ss)~{Potentials from Continuous Charge Distributions} to obtain an integral formula for the potential due to any continuous charge distribution, namely \begin{equation} V(\rr) = \int\limits_{\hbox{all charge}} {1\over 4\pi\epsilon_0} {\rho(\rrp)\, \delta\tau'\over|\rr-\rrp|} \end{equation} We also obtained a similar formula for the electric field in in~(ss)~{Electric Fields from Conti… text/html 2012-10-07T17:02:30-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/fieldint?rev=1349654550 Suppose that, instead of each of the students in the class being distinguishable, separate charges $q_i$, we instead pack many students together closely so that it makes more sense to idealize the charges in the room as a smoothed out charge density $\rho(\rr)$. What is the electric field in the room due to this charge density? text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/find?rev=1277055518 Since $\EE$ is the derivative of $V$, we should be able to recover $V$ from $\EE$ by integrating. This is in fact correct, as can be seen by recalling the Master formula: \begin{eqnarray*} dV = \grad{V}\cdot{d\rr} \end{eqnarray*} Integrating both sides yields the fundamental theorem for gradients, namely \begin{eqnarray*} V \bigg|_A^B = \Int_A^B\grad{V}\cdot{d\rr} \end{eqnarray*} This integral does not depend on the path of integration! If you move from one place to the other, the difference i… text/html 2019-02-07T15:09:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/gauss?rev=1549580940 Gauss's Law says that a geometric property of the electric field (namely the flux through a particular closed surface) is the same as the charge enclosed by that same surface (divided by $\epsilon_0$ in the system of units that we are using): \begin{eqnarray*} \oint_{\hbox{closed surface}} \EE \cdot d\AA = \frac{1}{\epsilon_0} \, Q_{\hbox{enclosed}} \end{eqnarray*} text/html 2012-12-31T20:55:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/gaussact?rev=1357016100 Choose one or more of the charge distributions given below: ($\alpha$ and $k$ are constants with appropriate dimensions.) \begin{enumerate}\item A positively charged (dielectric) spherical shell of inner radius $a$ and outer radius $b$ with a spherically symmetric internal charge density $\rho(\vec r)=\alpha\, r^3$. \item A positively charged (dielectric) spherical shell of inner radius $a$ and outer radius $b$ with a spherically symmetric internal charge density $\rho(\vec r)=\alpha\, e^{(kr)^… text/html 2019-02-07T15:35:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/gausshint?rev=1549582500 text/html 2019-02-07T15:07:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/gausssymmetry?rev=1549580820 In (ss)The Electric Field of a Uniform Disk, we found, by a round-about, complicated argument that the electric field due to an infinite, uniform plane of charge is give by \begin{eqnarray*} \EE(z) = \hbox{sgn}(z) \> \frac{\sigma}{2\epsilon_0}\,\zhat \end{eqnarray*} where the notation ${\rm sgn}(z)$ represents the sign of $z$. In cases like this, with very high symmetry, it is often simpler to use Gauss's law to find the electric field. text/html 2012-12-31T11:46:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/gradientact?rev=1356983160 The gradient of a function of three variables is a vector at each point in space. How can we graph such vector fields? How many different ways can you represent this information? After you have thought about these questions yourself, you can use the Maple worksheet [Visualizing the Gradient] or the Mathematica notebook [Visualizing the Gradient] to explore several different mechanisms for visualizing the gradient in two and three dimensions. text/html 2019-02-07T15:16:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/gradienthint?rev=1549581360 Figure 1:The gradient in 2 dimensions. In (ss)~{Activity: Using Technology to Visualize the Gradient}, you were asked to explore different ways of representing the gradient graphically. One combined representation is shown in Figure~1, showing both the gradient vector field and the level curves. text/html 2012-12-31T20:50:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/hillnavact?rev=1357015800 Suppose you are standing on a hill. You have a topographic map, which uses rectangular coordinates $(x,y)$ measured in miles. Your global positioning system says your present location is at one of the following points (pick one): A: $(1,4)$ ~~~~ B: $(4,-9)$ ~~~~ C: $(-4,9)$ ~~~~ D: $(1,-4)$ ~~~~ E: $(2,0)$ ~~~~ F: $(0,3)$ text/html 2019-02-07T15:22:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/hillnavhint?rev=1549581720 Figure~1:The topographic map of the hill Figuring out the location of the top of the hill, and its height, should have been easy. But drawing level curves --- the topo map for the hill, shown in Figure~1 --- is not so easy. Don't skip this step! Feel free to use a graphics program to generate these drawings, but it is important to develop the ability to translate between equations and topo maps. text/html 2015-08-16T08:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/intro?rev=1439738040 Idealizations are concepts like point charges, infinitely thin wires, massless rods, infinitely sharp barriers, etc. --- physics is full of them. They are enormously useful, but physicists need to understand how to deal with the real world, which is messier. To do this, physicists often exploit approximations. Thus, you will need to learn when and how to approximate and when and how to idealize. We'll use electrostatics as an example of how to use approximations. text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/lines?rev=1277055518 You probably already know the following facts about electric field lines: \begin{itemize}\item The density of field lines is proportional to the strength of the electric field in that area; \item Field lines only start at positive charges and end at negative charges; \item Field lines never cross. \end{itemize} These rules can all be explained using Gauss' Law, since the flux of the electric field can be interpreted as counting the number of field lines which cross the surface. text/html 2012-10-28T14:57:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/maxwell1?rev=1351461420 Recall that Gauss' Law says that \begin{eqnarray*} \Int_{\rm box} \EE \cdot d\AA = \frac{1}{\epsilon_0} \, Q_{\hbox{inside}} \end{eqnarray*} But the enclosed charge is just \begin{eqnarray*} Q_{\hbox{inside}} = \Int_{\rm box} \rho \> dV \end{eqnarray*} so we have \begin{eqnarray*} \Int_{\rm box} \EE \cdot d\AA = \frac{1}{\epsilon_0} \Int_{\rm box} \rho \> dV \end{eqnarray*} Applying the Divergence theorem to the left-hand side of this equation tells us that \begin{eqnarray*} \Int_{\hbox{inside… text/html 2012-12-31T11:53:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/maxwell2?rev=1356983580 Recall that Amp\`ere's Law says that the circulation of the magentic field around any imaginary closed loop (not around a physical loop, such as a wire) is proportional to the current enclosed by the loop, \begin{eqnarray*} \oint\limits_{\rm loop} \BB \cdot d\rr = \mu_0 \, I_{\rm enclosed} \end{eqnarray*} But the enclosed current is just the flux integral of the current density through any surface bounded by the loop, \begin{eqnarray*} I_{\rm enclosed} = \Int_{\rm surface} \JJ\cdot d\AA \end{e… text/html 2012-10-24T13:58:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/phflux?rev=1351112280 The most natural direction associated with a surface is the direction perpendicular to it. Therefore, the natural integral to compute over a surface adds up the normal component of $\EE$. This is called the flux of $\EE$ through the given surface (in the given direction): \begin{eqnarray*} \hbox{flux} = \int \EE \cdot \nn \, dA = \int \EE \cdot d\AA \end{eqnarray*} Be warned that some authors use two integral signs rather than one, and other letters, such as $S$ or $\sigma$, are often used in … text/html 2019-02-07T15:16:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/position?rev=1549581360 In the real world, if you want to find the distance between two objects, you can simply get out your trusty meter stick and measure it. In calculations, it is not quite so simple. First, you must figure out how to describe where the two objects are. To do this, you must choose an origin, and then you must locate the two objects by finding their position vectors $\rr$ and $\rrp$ with respect to that origin. See Figure~1. Figure 1: The position vectors for two objects and the (vector) di… text/html 2012-09-18T20:14:28-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/positioncurv?rev=1348024468 The position vector field is easier to write algebraically in rectangular coordinates than it is to think about: \begin{equation} \rr=x\,\ii+y\,\jj+z\,\kk \label{position} \end{equation} But what about in curvilinear coordinates? If you try to write the position vector $\rr(P)$ for a particular point $P$ in spherical coordinates, and you think of the tail of the position vector as ``attached'' to the origin, then you have a problem. It is not clear which $\rhat$, $\that$, and $\phat$ you shou… text/html 2012-10-07T16:31:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/potentialint?rev=1349652660 Suppose that, instead of each of the students in the class being distinguishable, separate charges $q_i$, we instead pack many students together closely so that it makes more sense to idealize the charges in the room as a smoothed out charge density $\rho(\rr)$. What is the electric potential in the room due to this charge density? text/html 2015-08-16T17:27:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/potentials?rev=1439771220 Recall that the electrostatic potential $V$ due to a point charge $q$ stationary at the origin is given by \begin{equation} V = \frac{1}{4\pi\epsilon_0} \> \frac{q}{r} \label{vpt} \end{equation} where $r$ is the distance from the point charge to the place at which the potential is being evaluated. Similarly, the gravitational potential $\Phi$ due to a point mass $m$ is given by \begin{equation} \Phi = -G \> \frac{m}{r} \end{equation} In the coming sections of the text, we will be discussing bot… text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/potentials2?rev=1277055518 We recalled the formula for the electric potential $V$ due to a single point charge. This can be thought of as a scalar field everywhere in space. The potential energy experienced by a test charge $q_{\rm test}$ is given by $q_{\rm test}V$; the test charge will move, if possible, so as to minimize the potential energy. This is very much like a 3-dimensional ``topo map'', with the test charge trying to ``roll downhill''. text/html 2019-02-07T15:22:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/pvector?rev=1549581720 The position vector is a unique and very peculiar beast. If you find yourself puzzling about the position vector, read this section. Otherwise, feel free to skip it. Consider a vector field, such as the electric field due to a point charge. It consists of a vector at each point in space, and is best represented by a vector graph where the tail of each vector is ``attached'' to the point in space at which the vector field is evaluated. text/html 2012-09-18T00:23:50-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ring?rev=1347953030 Find the electrostatic potential for a uniformly charged ring with total charge $Q$ and radius $R$. text/html 2012-10-28T14:54:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ringhint?rev=1351461240 In (ss)~{Activity: A Uniformly Charged Ring}, you will have found an integral expression for the electrostatic potential due to a uniform ring of charge. Some things you may have needed to pay attention to are: \begin{enumerate}\item Draw a picture! \item How thick is the ring? What shape is its cross section? Should it be described by a volume charge density? In the absence of further information, the only thing you can do is idealize the ring as a line charge. \item In equation~(1) o… text/html 2012-10-07T15:36:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ringlimit?rev=1349649360 Determine the power series exansions that represent the electrostatic potential due to the charged ring, both on axis and in the plane of the ring, and near to and far from the ring. text/html 2019-02-07T15:15:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/ringlimithint?rev=1549581300 Figure 1:The level curves of $V$, shown in a vertical plane with the $z$-axis at the left. The integral that you found in (ss)~{Activity: A Uniformly Charged Ring} is, in general, an elliptic integral. It cannot be evaluated in closed form other than to say that its value is the value of that integral. You can look up the value in a table (the old-fashioned way) or get a computer algebra package to compute the value for you (the modern way), but in either case, you do not get an answer whic… text/html 2015-08-16T09:11:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/scalarfields?rev=1439741460 The electrostatic potentials, $V$, and the gravitational potential, $\Phi$, are examples of scalar fields. A scalar field is any scalar-valued physical quantity (i.e.~a number with units attached) at every point in space. It may be useful to think of the temperature in a room, $T$, as your ideal of a scalar field. text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/second2?rev=1277055518 Since the electric field is (minus) the gradient of the scalar potential, it is conservative. But since \begin{eqnarray*} \grad\times\grad V = \zero \end{eqnarray*} for any function $V$, we can rewrite this as \begin{eqnarray*} \grad\times\EE = \zero \end{eqnarray*} which is the differential form of (the electrostatic version of) Faraday's Law, and another of Maxwell's Equations. text/html 2012-10-28T14:29:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/superposition?rev=1351459740 Suppose each of the $N$ students in the class is a separate charge. What is the electrostatic potential in the room due to these charges? The electrostatic potential due to a single charge is \begin{equation} V = \frac{1}{4\pi\epsilon_0} \> \frac{q}{r} \end{equation} where the $r$ in the denominator is the distance between the source and the observation point. text/html 2015-08-18T17:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/totalcharge?rev=1439943240 Suppose that a spherical shell $B$ has inner radius $a$ and outer radius $b$ with a spherically symmetric internal charge density $\rho(r)=10kr^2$. What is the total charge? Suppose instead that the shell is cylindrical, with the same inner and outer radii, but with a cylindrically symmetric internal charge density $\rho(r)=10kr^2$. What is the total charge? text/html 2015-08-18T19:11:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/totalchargehint?rev=1439950260 Recall that the volume element in spherical coordinates is given by $dV=r^2\sin\theta\,dr\,d\theta\,d\phi$. If the charge density on the spherical shell is $\rho(r)=10kr^2$ (in spherical coordinates), then the total charge on the shell is given by \begin{eqnarray} q_{\hbox{tot}} &=& \int_B\rho\,dV \nonumber\\ &=& \int_0^{2\pi} \int_0^\pi \int_a^b 10k\,r^2 r^2\sin\theta\,dr\,d\theta\,d\phi \nonumber\\ &=& 8k\pi\, (b^5-a^5) \end{eqnarray} text/html 2019-02-07T15:19:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/triangle?rev=1549581540 Starting with the vector potential for a line current, we used the superposition principle in~(ss)~{The Biot-Savart Law} to obtain an integral formula for the vector potential due to any current distribution, namely \begin{equation} \AA(\rr) = {\mu_0\over 4\pi} \int {\JJ(\rrp)\,d\tau'\over|\rr-\rrp|} \end{equation} as well as a similar formula for the magnetic field, the Biot-Savart Law, namely \begin{equation} \BB(\rr) = {\mu_0\over 4\pi} \int {\JJ(\rrp)\times(\rr-\rrp)\,d\tau'\over|\rr-\rrp… text/html 2012-10-07T20:15:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/veufact?rev=1349666100 Consider the four quantities $V$ (electrostatic potential), $\EE$ (electric field), $U$ (potential energy), and $\FF$ (force). Make a concept map showing the relationship between these quantities. That is, place these four quantities at the corners of a square, and draw arrows between them, indicating for each one what relationship it represents between these quantities. text/html 2012-10-07T20:20:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/veufhint?rev=1349666400 Here are some ideas that may improve your understanding of the VEUF activity. \begin{itemize}\item This is a good place to review your understanding of energy and force from previous coursework in physics. \item Don't forget the minus sign in the relationships --- both directions --- between potential and field. \item Note the parallels between the relationships potential/field and energy/force, and also between field/force and potential/energy. \item Make sure you understand both the mathematic… text/html 2019-02-07T15:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/visv?rev=1549581240 If a scalar field such as temperature or electrostatic potential or gravitational potential represents a scalar-valued physical quantity at each point in space, then, since space is three-dimensional, a scalar field is effectively a function of three variables. How can we graph such functions? For functions of one or two variables, if we use one or two dimensions to represent the domain of the scalar field, we have at least one physical dimension left to represent the value of the function. Th… text/html 2019-02-07T15:12:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/visvhint?rev=1549581120 In (ss)~{Activity: Using Technology to Visualize Potentials}, you were asked to explore different ways of graphically representing a scalar field such as the electrostatic potential in three dimensions. Several standard representations that are used by scientists are show below. text/html 2019-02-07T15:06:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/vlinefinite?rev=1549580760 As an example of finding the potential due to a continuous charge source, let's calculate the potential the distance $r$ from the center of a uniform line segment of charge with total length $2L$. We will idealize the line segment as infinitely thin and describe it by the constant linear charge density $\lambda$. Because we are chopping a one-dimensional source into little lengths, $d\tau$ reduces to $|d\rr|$. \begin{equation} V(\rr)=\frac{1}{4\pi\epsilon_0}\int \frac{\lambda |d\rr|}{|\rr-\rrp… text/html 2012-12-31T11:50:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/vlineinf?rev=1356983400 In (ss)~{Potential due to a Finite Line of Charge}, we found the electrostatic potential due to a finite line of charge. The answer \begin{eqnarray} V(r,0,0) &=& \frac{\lambda}{4\pi\epsilon_0} \ln\left(\frac{L + \sqrt{r^2+L^2}}{-L + \sqrt{r^2+L^2}}\right) \label{vfiniteline2} \end{eqnarray} was an unilluminating, complicated expression involving the logarithm of a fraction. Suppose, however, that the voltmeter probe were placed quite close to the charge. Then, to a fairly good approximation,… text/html 2015-08-16T09:06:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/voltmeters?rev=1439741160 In principle, a voltmeter measures the electrostatic potential, the physical quantity that we will be talking about in these sections of the book. But you cannot use a voltmeter to measure the electrostatic potential due to a collection of charges that are just sitting around in the lab, since that is not the way that voltmeters actually work in practice. Normally, you attach the two probes of a voltmeter to different places in a circuit. The voltmeter contains a really large resistor and mea… text/html 2012-10-08T11:33:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/vpoints?rev=1349721180 To check your understanding of this chapter, you should complete at least the first two cases from the following activity. (The mathematical manipulations in all of the cases are essentially the same, although different cases will teach you different things about the physics.) A discussion of some of the most important observations you may make while completing this activity is contained in the following section. Also included in the next section are the final answers to all of the problems i… text/html 2012-10-08T11:44:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/vpoints2?rev=1349721840 Start by writing down a formula for the electrostatic potential $V(\rr)=V(x,y,z)$ everywhere in space due to a single point charge that is not located at the origin. You should then find the electrostatic potential everywhere in space for the following two cases: \begin{enumerate}\item Two charges $+Q$ and $+Q$ are placed on a line at $z=+D$ and $z=-D$, respectively. \item Two charges $+Q$ and $-Q$ are placed on a line at $z=+D$ and $z=-D$, respectively. \end{enumerate} text/html 2012-10-08T11:50:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/vpoints2hint?rev=1349722200 You may be tempted to use the iconic equation for the electrostatic potential to give an answer along the lines of $$V=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)$$ But this expression doesn't provide enough information to describe $r_1$ and $r_2$, a good example of the need to ``unpack'' the iconic equation. text/html 2012-10-28T14:14:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/vpointshint?rev=1351458840 (ss)~{Activity: Power Series for Dipoles} combines both the math (power series) and physics (electrostatic potential) we've been discussing. All of the cases can be approximated by exploiting one of the common power series that it was suggested that you memorize --- there's no need to compute the coefficients using differentiation. But it's important to understand the approximation process: Make sure you are expanding in terms of something small! This requires rewriting each potential, typical… text/html 2012-10-10T20:03:00-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/vpot?rev=1349924580 Recall the use of the superposition principle to find the potential due to any charge distribution, starting from the electric potential due to a point charge, namely \begin{eqnarray*} V(\rr) = {1\over 4\pi\epsilon_0} {q\over|\rr-\rr'|} \end{eqnarray*} resulting in \begin{eqnarray*} V(\rr) = {1\over 4\pi\epsilon_0} \int {\rho(\rrp)\,d\tau'\over|\rr-\rrp|} \end{eqnarray*} text/html 2010-06-20T10:38:38-08:00 http://sites.science.oregonstate.edu/BridgeBook/book/physcontent/wire?rev=1277055518 Consider the magnetic field of a straight wire along the $z$-axis carrying a steady current $\II=I\,\zhat$. The Biot-Savart Law says that \begin{eqnarray*} \BB = - {\mu_0 I\over 4\pi\epsilon_0} \Int_{-L}^{L} {(\rr-\rrp)\times d\rrp\over|\rr-\rrp|^3} \end{eqnarray*} and we have \begin{eqnarray*} \rr &=& r\,\rhat + z\,\zhat \\ \rrp &=& z'\,\zhat \end{eqnarray*} so that \begin{eqnarray*} - (\rr-\rrp)\times d\rr &=& - \bigl(r\,\rhat + (z-z')\,\zhat\bigr) \times dz'\,\zhat \\ &=& r\,dz'\,\phat \e…