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potential equation physics
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potential equation physics
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potential equation physics
potential equation physics
In the usual case, $V$ would depend on $x$, $y$, and $z$, and the differential equation must be integrated to reveal the simultaneous dependence on these three variables. Common types of potential energy include the gravitational potential energy of an object, the elastic potential energy of an extended spring, and the electric potential energy of an electric charge in an electric field. V is the electric potential measured by volts (V). (Negative integers are excluded, because $\alpha$ must be positive.) Evaluating the potential of Eq. This problem could have been solved more formally by inverting the spherical-harmonic series along the lines described in Sec.8.3. The answer is obvious if \( r \) is constant: we get a circular orbit. \end{aligned} It would be great to have a 15m chat to discuss a personalised plan and answer any questions. You may find the identity \[ \frac{d}{du} \text{arctan}(u) = \frac{1}{1 + u^2} \] useful to work through this problem. This gives, \begin{equation} V(r,\theta,\phi) = V_0 \biggl[ \frac{1}{3} - \frac{1}{6} (r/R)^2 (3\cos^2\theta - 1) + \frac{1}{2} (r/R)^2 \sin^2\theta \cos(2\phi) \Biggr], \tag{10.86} \end{equation}. Exercise 10.6: In case you are skeptical that the method described above leads to the correct solution, verify that the potential of Eq. (10.29) in the case of the parallel plates. Last time, we continued our study of the two-body central force problem, and found that conservation of angular momentum forces all the motion to happen in a plane, so that we can use the two polar coordinates \( (r, \phi) \) to describe this system. The form of the differential equation suggests the use of $r^\alpha$ as a trial solution, where $\alpha$ is a constant. The standard cell potential is the potential difference between the cathode and anode. We wish to find the electrostatic potential everywhere between the two side plates and above the bottom plate. g denotes the acceleration due to gravity. Notice that the potential is discontinuous at $(x,y) = (0,0)$ and $(x,y) = (L,0)$; this can be achieved by inserting a thin layer of insulating material between the bottom and side plates. This reveals the existence of a rotational symmetry: nothing changes physically as we rotate around the $z$-axis, and it follows that the potential cannot depend on $\phi$. for the final solution to our boundary-value problem. (Boas Chapter 12, Section 2, Problem 3) Consider the problem of the parallel plates, as in Sec.10.3, but assume now that the bottom plate is maintained at $V = V_0 \cos x$. Elastic Potential Energy: The energycontained in compressible or stretchable items such as elastic bands, trampolines, and bungee cords is called elastic potential energy. where $A_{nm}$ and $B_{nm}$ are arbitrary expansion coefficients. Energy is always conserved. Also, the coefficient of friction between the two surfaces is small = 0.05. \begin{aligned} and a short calculation yields $b_n = 2V_0(1-\cos n\pi)/(n \pi)$. The electric potential V of a point charge is given by. The reason is that $x$, $y$, and $z$ are all independent variables. Consider the example above where a person applies a force to move a crate. The coefficients would decrease even faster if the potential didn't present discontinuities at $(x,y) = (0,0)$ and $(x,y) = (L,0)$. Techniques to invert Legendre series were described back in Sec.8.2, and Eq. Find the potential in the region described by $0 < x < 1$ and $0 < y < 1$. fuel and explosives have Chemical PE a coiled spring or a drawn bow also have PE due to their state Kinetic energy (KE) is energy of motion A moving car has a lot of kinetic energy From PE to KE To give it an interpretation, let us write $A_0 := -q/(4\pi \epsilon_0 R)$, and therefore express $A_0$ in terms of another constant $q$ with the dimension of charge. The potential is, \[ . While this expression doesn't seem to involve spherical harmonics, in fact they are there in disguise. \tag{10.12} \end{equation}, \begin{equation} Y(y) = e^{\pm i\beta x}, \tag{10.13} \end{equation}, \begin{equation} Y(y) = \left\{ \begin{array}{l} \cos(\beta y) \\ \sin(\beta y) \end{array} \right. The equation is PEspring = 0.5 k x2 where k = spring constant Your solution will be expressed as a Fourier series. Evaluating the potential of Eq. \mu \ddot{r} = -\frac{dU_{\textrm{eff}}}{dr}. We know that $V$ must vanish at $y = \infty$, and this implies that the solution cannot include a factor $e^{\alpha y}$, which grows to infinity. Our result for $\hat{A}_{nm}$ implies that, \begin{equation} 2 A_{nm} = \frac{16 V_0}{\pi^2} \frac{1}{nm} \frac{1}{ \sinh\bigl( \sqrt{n^2+m^2}\, \pi b/a \bigr)}, \tag{10.40} \end{equation}, \begin{equation} V(x,y,z) = \frac{16 V_0}{\pi^2} \sum_{n=1,3,\cdots}^\infty \sum_{m=1,3,\cdots}^\infty \frac{1}{nm} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr) \frac{ \sinh\bigl( \sqrt{n^2+m^2}\, \pi z/a \bigr) }{ \sinh\bigl( \sqrt{n^2+m^2}\, \pi b/a \bigr) } \tag{10.41} \end{equation}. a) What are the factorized solutions for the two-dimensional Laplace equation $\nabla^2 V = 0$ expressed in polar coordinates $s$ and $\phi$? We wish to calculate the electric potential $V$, under the assumption that the conductor is maintained at $V=0$ during the immersion. To represent this constant field at large distances we need a potential that behaves as $V \sim -E z$, or, \begin{equation} V \sim - E\, r \cos\theta, \qquad r \to \infty. The direction of the moment is normal to the plane of the loop . We wish to find $V$ everywhere within the box. Electric Potential Formula. These techniques rest on what was covered in previous chapters. With this restriction on $\mu$, Eq. (10.18) and (10.19), which we write in the hybrid form, \begin{equation} V_{\alpha,\beta}(x,y,z) = \left\{ \begin{array}{l} \cos(\alpha x) \\ \sin(\alpha x) \end{array} \right\} \left\{ \begin{array}{l} e^{i\beta y} \\ e^{-i\beta y} \end{array} \right\} \left\{ \begin{array}{l} e^{\sqrt{\alpha^2+\beta^2}\, z} \\ e^{-\sqrt{\alpha^2+\beta^2}\, z} \end{array} \right\}. for the unique solution to the boundary-value problem, expressed as an infinite sum of products of Bessel and exponential functions. (k - \mu \dot{\phi}^2) r = k\ell \Rightarrow r = \frac{\ell}{1 - (\mu/k) \dot{\phi}^2}. We begin in this chapter with one of the most ubiquitous equations of mathematical physics, Laplace's equation, \begin{equation} \nabla^2 V = 0. U_{\textrm{eff}} = \frac{L_z^2}{2\mu r^2} This is not an initial condition, it's an equilibrium solution. \begin{aligned} It can be shown that for a generic value of $\mu$, the solutions to Eq. You will get the wrong answer! If we begin by considering \( \dot{\phi} \) relatively small, so the denominator remains between 0 and 1, then this expression makes sense: as \( \dot{\phi} \) increases, the equilibrium value of \( r \) is pushed out towards values larger than \( \ell \), where it would be if there was no rotational motion. Connect with a tutor from a university of your choice in minutes. This can be achieved by demanding that $kR$ be a zero of the Bessel function, so that $k = \alpha_{0p}/R$, where, in the notation introduced in Sec.5.3, $\alpha_{0p}$ is the $p^{\rm th}$ zero of the zeroth Bessel function. the energy due to position of a quantity in a field. With $u = \cos\theta$, we are speaking of functions that become infinite at $\theta = \pi$, just like the $Y$ of Eq.(10.73). This is the statement of the superposition principle, and it shall form an integral part of our strategy to find the unique solution to Laplace's equation with suitable boundary conditions. Before the immersion the field was truly constant, but the arrival of the conductor distorts the electric field, because of the induced charge distribution on the surface; the final field is not quite uniform. which we shall insert within Laplace's equation. The electric potential due to a point charge q at a distance of r from that charge is mentioned by: V = q/(4 0 r) In this equation, 0 is the permittivity of free space. U_{\textrm{eff}}(r) = U(r) + \frac{L_z^2}{2\mu r^2}, In this section we suppose that the boundary surfaces are cylinders, and consider solving Laplace's equation using the cylindrical coordinates $(s,\phi,z)$. Find the GPE of an object of mass 10 kg raised 20 m above the ground. JOHN M. COWLEY, in Diffraction Physics (Third Edition), 1995. We shall need the curvilinear coordinates of Chapter 1, the special functions of Chapters 2, 3, 4, 5, and 6, and the expansion in orthogonal functions of Chapters 7, 8, and 9. The list could go on. While Cartesian coordinates are a good choice when Laplace's equation is supplied with boundary conditions on planar surfaces, other coordinates can be better suited to other geometries. 1-519-824-4120 x 52261 \begin{aligned} We will expand on that discussion here as we make an effort to associate the motion characteristics described above with the concepts of kinetic energy, potential energy and total mechanical energy.. Evaluate the potential at $s = \frac{1}{2} R$ and $z = R$. Imagine to be in 2 dimensions and you have to find the potential generated by 4 point-charges of equal charge located at the four corners of a square. Potential energy = (charge of the particle) (electric potential) U = q V U = qV Derivation of the Electric Potential Formula U = refers to the potential energy of the object in unit Joules (J) In general, to have motion with fixed \( r \), the system must be stuck exactly at a single turning point, which means the \( E \) line must be just touching the effective potential curve; this will happen at the minimum possible \( E \). We found that it is a dipole field, with the dipole moment given by = IA, where I is the current and A is the area of the loop. The external work done per unit charge is equal to the change in potential of a point charge. \tag{10.45} \end{equation}. It's easy to sketch what the shape of the potential will look like for various choices of \( L_z \): Why is plotting \( U_{\textrm{eff}} \) useful? (10.32) at $z=0$ yields, \begin{equation} 0 = \sum_{n=1}^\infty \sum_{m=1}^\infty \bigl( A_{nm} + B_{nm} \bigr) \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr), \tag{10.33} \end{equation}, and we find that the expansion coefficients must be related by $B_{nm} = -A_{nm}$. Equation (10.8) states that a function of $x$ only is equal to the sum of a function of $y$ only and a function of $z$ only. Furthermore, the mass of the block of ice is 250 kg. However, that's not really the case. and members of the basis can now be labelled with the integer $n$. \tag{10.1} \end{equation}. \tag{10.3} \end{equation}. (4.44), the differential equation satisfied by the spherical harmonics. The upper hemisphere is maintained at $V = V_0$, while the lower hemisphere is maintained at $V = -V_0$. We could factorize $Y(\theta,\phi)$ further by writing it as $\Theta(\theta) \Phi(\phi)$, but this shall not be necessary. After-lecture aside: what is the equilibrium value of \( r \) here? We cannot expect all solutions to Laplace's equation to be of this simple, factorized form; the vast majority are not. Remember this is completely equivalent to the 1-D problem of finding the motion of \( r \) in such a potential. U_{\textrm{eff}} = -\frac{GM_s m}{r} + \frac{L_z^2}{2\mu r^2}. The factorized solutions become, \begin{equation} V_p(s,z) = J_0(\alpha_{0p} s/R)\, e^{-\alpha_{0p} z/R}, \tag{10.62} \end{equation}. There is a special equation for springs that relates the amount of elastic potential energy to the amount of stretch (or compression) and the spring constant. CUPE 3913 which is exactly the total energy from our original two-dimensional Lagrangian. Changes in membrane potential elicit action potentials and give cells the ability to send messages around the body. Gryph Mail Because of all this freedom, and because $\alpha$ and $\beta$ are arbitrary parameters, we are quite far from having a unique solution to Laplace's equation. Spherical boundaries call for spherical coordinates, and in this section we take on the task of solving Laplace's equation in these coordinates. A potential term is just an ''interaction term'' for scattering experiments so we can add this in in the case of the continuity of eignevalues and [a] solution can be found using the Lippmann-Schwinger equation, just as an example. The idea is to select the blocks that best suit the given problem, and to superpose them so as to satisfy the boundary conditions. Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. The elastic potential energy stored can be calculated using the equation: elastic potential energy = 0.5 spring constant (extension . The more an objects ability to stretch, the greater its elastic potential energy. So both B and D are true. (10.72) becomes, \begin{equation} r^2 \frac{d^2 R}{dr^2} + 2r \frac{dR}{dr} = \ell(\ell+1) R, \tag{10.77} \end{equation}. To achieve this we must require that $\sin(\alpha x) = 0$ at $x = L$, so that $\alpha L$ must be a multiple of $\pi$. For the spring, we find that, \[ Suppose that $V_1$, $V_2$, $V_3$, and so on, are all solutions to Laplace's equation, so that $\nabla^2 V_j = 0$. The factorized solutions to Laplace's equation in spherical coordinates are therefore, \begin{equation} V^m_\ell(r,\theta,\phi) = \left\{ \begin{array}{l} r^\ell \\ r^{-(\ell+1)} \end{array} \right\} Y^m_\ell(\theta,\phi), \tag{10.78} \end{equation}, and they are labelled by the integers $\ell$ and $m$ that enter the specification of the spherical harmonics. If there is no angular momentum, then we only have the second term, which is just the force acting in the radial direction, \( F = -\nabla U \). The only possible way for \( U_{\textrm{eff}} \) to be zero is if both \( r=\ell \) and \( L_z = 0 \) - but this point is precisely the minimum value of \( E \) given those conditions! Physics (Single Science) PSHE and Citizenship . Then show that the expansion coefficients for the constant function $V_0$ are given explicitly by $\hat{A}_{nm} = 16 V_0/(nm \pi^2)$ when $n$ and $m$ are both odd. A generalization to the associated Legendre equation, \begin{equation} (1-u^2) f'' - 2u f' + \biggl[ \lambda(\lambda+1) - \frac{m^2}{1-u^2} \biggr] f = 0, \tag{10.75} \end{equation}. The final solution to our boundary-value problem will be a superposition of these basis solutions. In this case this is the best we can do; there is no simpler expression for the solution, analogous to Eq. It is important to understand that the superposition principle applies to any number of solutions $V_j$; this number could be (and will be) infinite. The third plate is infinite in the $z$-direction only, and it is maintained at $V = V_0$. This is the case here also, as suggested by the fact that the coefficients decrease as $1/n$ with increasing $n$. Work entails the use of a force to shift an object. where $\alpha$ and $\beta$ are arbitrary parameters. Because all plates are infinite in the $z$-direction, nothing changes physically as we move in that direction, and the system is therefore symmetric with respect to translations in the $z$-direction. The boundary conditions and Eq. Again we can go back and forth between the complex exponentials and the trigonometric functions, and the sign in front of $\beta^2$ can be altered by letting $\beta \to i\beta$. But if \( r \) is varying with time, just knowing the minimum and maximum possible values doesn't tell us much about what happens in the middle: Next time: some review, and then we find the equation describing the orbital shape. In particular if I choose the origin of the cartesian coordinates at the center of the square I get (the side . \]. (10.73) becomes <\p> \begin{equation} \frac{1}{\sin\theta} \frac{\partial}{\partial \theta} \biggl( \sin\theta \frac{\partial Y}{\partial \theta} \biggr) + \frac{1}{\sin^2\theta} \frac{\partial^2 V}{\partial \phi^2} = -\ell(\ell+1) Y, \tag{10.76} \end{equation}, which is precisely the differential equation for the spherical harmonics. . \tag{10.9} \end{equation}, The solutions to this ordinary differential equation are, \begin{equation} X(x) = e^{\pm i\alpha x}, \tag{10.10} \end{equation}, \begin{equation} X(x) = \left\{ \begin{array}{l} \cos(\alpha x) \\ \sin(\alpha x) \end{array} \right. = refers to the coefficient of friction = refers to the normal force acting on the object Solved Example on Friction Formula Example 1 Assume a large block of ice is being pulled across a frozen lake. Let's instead turn to the more interesting case of gravitational interaction. (10.63) gives, \begin{equation} V(s,z) = 2 V_0 \sum_{p=1}^\infty \frac{1}{\alpha_{0p} J_1(\alpha_{0p})} J_0(\alpha_{0p} s/R)\, e^{-\alpha_{0p} z/R} \tag{10.68} \end{equation}. We know that \( r \) changes as the particle moves along the line; it decreases to a minimum as \( m_1 \) approaches \( m_2 \), and then starts to increase again. (10.29) satisfies $\nabla^2 V = 0$ and the boundary conditions specified at the beginning of the section. makes the situation even worse. (8.24) informs us that the coefficients are given by, \begin{equation} c_p = \frac{2}{\bigl[ J_1(\alpha_{0p}) \bigr]^2} \int_0^1 V_0 J_0(\alpha_{0p} u)\, u\, du, \tag{10.65} \end{equation}, \begin{equation} c_p = \frac{2V_0}{\bigl[ \alpha_{0p} J_1(\alpha_{0p}) \bigr]^2} \int_0^{\alpha_{0p}} v J_0(v)\, dv \tag{10.66} \end{equation}, by introducing the new integration variable $v := \alpha_{0p} u$. (10.67) within Eq. Enter Displacement. (5.26) reveals that this differential equation is none other than Bessel's equation. At this stage we have obtained that the solution to the boundary-value problem must be built from, \begin{equation} V_k(s,z) = J_0(ks)\, e^{-kz}, \tag{10.61} \end{equation}, The potential is required to go to zero at $s = R$. We write this as, \begin{align} V(x,y,z) &= \sum_{n=1}^\infty \sum_{m=1}^\infty \biggl[ A_{nm} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr) e^{\sqrt{n^2+m^2}\, \pi z/a} \nonumber \\ & \quad \mbox{} + B_{nm} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr) e^{-\sqrt{n^2+m^2}\, \pi z/a} \biggr], \tag{10.32} \end{align}. The superposition principle follows directly from the fact that Laplace's equation is linear in the potential $V$. Find the GPE of an object of mass 2 kg raised 6 m above the ground. Potential theory. The wall of the pipe is maintained at $V = 0$, and its base is maintained at $V = V_0 (s/R) \sin\phi$. The same condition implies that $B_0 = -A_0 R$, and we now have, \begin{equation} V = A_0 (1 - R/r) - E (r - R^3/r^2) \cos\theta \tag{10.90} \end{equation}, Notice that $V$ depends on an unknown constant $A_0$ that is not determined by the boundary and asymptotic conditions. I've drawn three energy levels on the potential plot. It is the energy generated as a result of an object's location in relation to other items. Collecting results, we find that the factorized solutions to Laplace's equation in Cartesian coordinates are of the form, \begin{equation} V_{\alpha,\beta}(x,y,z) = \left\{ \begin{array}{l} e^{i\alpha x} \\ e^{-i\alpha x} \end{array} \right\} \left\{ \begin{array}{l} e^{i\beta y} \\ e^{-i\beta y} \end{array} \right\} \left\{ \begin{array}{l} e^{\sqrt{\alpha^2+\beta^2}\, z} \\ e^{-\sqrt{\alpha^2+\beta^2}\, z} \end{array} \right\}, \tag{10.18} \end{equation}. Respiration (cellular respiration, that is, not pulmonary respiration or breathing as it is better known) is chemically identical to combustion, but it takes place at a much slower rate. Various forms of energy are studied in physics. Potential difference is also known as voltage. This physics video tutorial provides a basic introduction into kinetic energy and potential energy. (10.37) are given by Eq.(10.39). The potential must also vanish at $x = 0$, and this rules out the presence of a factor $\cos(\alpha x)$. In mathematics and mathematical physics, potential theory is the study of harmonic functions . Find the potential inside the pipe. This idea will be made concrete in the following sections. The charge is given in terms of micro-Coulombs (C): 1.0 C = 1.0 x 10 -6 C. The charge needs to be converted to the correct units before solving the equation: VB = 300 V - 100 V. We recognize the $-E\, r\cos\theta$ contribution to the potential, which gives rise to the constant field at large distances, but we also see a correction proportional to $R^3/r^3$, which comes from the distribution of surface charge on the conductor. This equation is encountered in electrostatics, where $V$ is the electric potential, related to the electric field by $\boldsymbol{E} = -\boldsymbol{\nabla} V$; it is a direct consequence of Gauss's law, $\boldsymbol{\nabla} \cdot \boldsymbol{E} = \rho/\epsilon$, in the absence of a charge density. For example, we need to dig further to answer the simple question: what is the shape of an orbit with a given total energy \( E \)? However, remember that \( \dot{\phi} \) is not arbitrary here! where $c_\ell \propto A^0_\ell$ are the expansion coefficients, and where the factor of $R^{-\ell}$ was inserted for convenience. The Laplacian operator was expressed in these coordinates back in Eq. The solution to this problem will be of the form of Eq. \end{aligned} With $\Phi = e^{\pm im\phi}$, this can be achieved if and only if $2m\pi$ is a multiple of $2\pi$, and this means that $m$ must be an integer, $m = 0, 1, 2, 3, \cdots$. For the gravitational force, the formula is: W = mgh = mgh Where, m is the mass in kilograms g is the acceleration due to gravity h is the height in meters Potential Energy Unit Gravitational potential energy has the same units as kinetic energy: kg m2 / s2 [email protected], College of Engineering & Physical Sciences, College of Social & Applied Human Sciences, Gordon S. Lang School of Business & Economics, Government Relations & Community Engagement. An object which is not raised above the ground will have a height of zero and therefore zero potential energy. (10.45) can be re-expressed as, \begin{equation} m^2 = \frac{s}{S} \frac{d}{ds} \biggl( s \frac{dS}{ds} \biggr) + \frac{s^2}{Z} \frac{d^2 Z}{dz^2}, \tag{10.49} \end{equation}, \begin{equation} -\frac{1}{Z} \frac{d^2 Z}{dz^2} = -\frac{m^2}{s^2} + \frac{1}{sS} \frac{d}{ds} \biggl( s \frac{dS}{ds} \biggr). (10.85) and multiply it by the appropriate factor of $(r/R)^\ell$. Graduate Calendar As a first example of a boundary-value problem formulated in spherical coordinates, we examine a system consisting of two conducting hemispheres of radius $R$ joined together at the equator (see Fig.10.7). This is known as gravitational potential energy. The first is that its solutions are unique once a suitable number of boundary conditions are specified. \]. The answer is that there was no good reason; we made an arbitrary choice, admittedly with an ulterior motive in mind (you'll see). And because the potential should not blow up at the centre of the sphere, we must also kill all $r^{-(\ell+1)}$ terms by setting $B^m_\ell = 0$. Because of the unequal concentrations of ions across a membrane, the membrane has an electrical charge. Each side of the box is maintained at $V=0$, except for the top side, which is maintained at $V=V_0$. The $q/(4\pi \epsilon_0 r)$ term in the potential comes with a correction proportional to $r/R$, and this represents an irrelevant constant. Consider an arbitrary, but finite, charge distribution, ( ), such as that illustrated in Figure (2.1.1). Exercise 10.1: Verify these results for the expansion coefficients $b_n$. The solution is now determined up to the expansion coefficients $A_{nm}$. Proceeding in a similar manner for the function of $y$, we write $g(y) = -\beta^2 = \text{constant}$, or, \begin{equation} \frac{1}{Y} \frac{d^2 Y}{dy^2} = -\beta^2. ), One more note: despite my warning above about Lagrangian substitution, remember that from the perspective of the one-dimensional problem, it's perfectly consistent to treat \( U_{\rm eff} \) as an effective potential energy. The more massive an object is, the greater its gravitational potential energy. (1.75), and we have, \begin{equation} 0 = \nabla^2 V = \frac{1}{s} \frac{\partial}{\partial s} \biggl( s \frac{\partial V}{\partial s} \biggr) + \frac{1}{s^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} \tag{10.42} \end{equation}, \begin{equation} V = S(s) \Phi(\phi) Z(z), \tag{10.43} \end{equation}, the product of functions of $s$, $\phi$, and $z$. 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In the usual case, $V$ would depend on $x$, $y$, and $z$, and the differential equation must be integrated to reveal the simultaneous dependence on these three variables. Common types of potential energy include the gravitational potential energy of an object, the elastic potential energy of an extended spring, and the electric potential energy of an electric charge in an electric field. V is the electric potential measured by volts (V). (Negative integers are excluded, because $\alpha$ must be positive.) Evaluating the potential of Eq. This problem could have been solved more formally by inverting the spherical-harmonic series along the lines described in Sec.8.3. The answer is obvious if \( r \) is constant: we get a circular orbit. \end{aligned} It would be great to have a 15m chat to discuss a personalised plan and answer any questions. You may find the identity \[ \frac{d}{du} \text{arctan}(u) = \frac{1}{1 + u^2} \] useful to work through this problem. This gives, \begin{equation} V(r,\theta,\phi) = V_0 \biggl[ \frac{1}{3} - \frac{1}{6} (r/R)^2 (3\cos^2\theta - 1) + \frac{1}{2} (r/R)^2 \sin^2\theta \cos(2\phi) \Biggr], \tag{10.86} \end{equation}. Exercise 10.6: In case you are skeptical that the method described above leads to the correct solution, verify that the potential of Eq. (10.29) in the case of the parallel plates. Last time, we continued our study of the two-body central force problem, and found that conservation of angular momentum forces all the motion to happen in a plane, so that we can use the two polar coordinates \( (r, \phi) \) to describe this system. The form of the differential equation suggests the use of $r^\alpha$ as a trial solution, where $\alpha$ is a constant. The standard cell potential is the potential difference between the cathode and anode. We wish to find the electrostatic potential everywhere between the two side plates and above the bottom plate. g denotes the acceleration due to gravity. Notice that the potential is discontinuous at $(x,y) = (0,0)$ and $(x,y) = (L,0)$; this can be achieved by inserting a thin layer of insulating material between the bottom and side plates. This reveals the existence of a rotational symmetry: nothing changes physically as we rotate around the $z$-axis, and it follows that the potential cannot depend on $\phi$. for the final solution to our boundary-value problem. (Boas Chapter 12, Section 2, Problem 3) Consider the problem of the parallel plates, as in Sec.10.3, but assume now that the bottom plate is maintained at $V = V_0 \cos x$. Elastic Potential Energy: The energycontained in compressible or stretchable items such as elastic bands, trampolines, and bungee cords is called elastic potential energy. where $A_{nm}$ and $B_{nm}$ are arbitrary expansion coefficients. Energy is always conserved. Also, the coefficient of friction between the two surfaces is small = 0.05. \begin{aligned} and a short calculation yields $b_n = 2V_0(1-\cos n\pi)/(n \pi)$. The electric potential V of a point charge is given by. The reason is that $x$, $y$, and $z$ are all independent variables. Consider the example above where a person applies a force to move a crate. The coefficients would decrease even faster if the potential didn't present discontinuities at $(x,y) = (0,0)$ and $(x,y) = (L,0)$. Techniques to invert Legendre series were described back in Sec.8.2, and Eq. Find the potential in the region described by $0 < x < 1$ and $0 < y < 1$. fuel and explosives have Chemical PE a coiled spring or a drawn bow also have PE due to their state Kinetic energy (KE) is energy of motion A moving car has a lot of kinetic energy From PE to KE To give it an interpretation, let us write $A_0 := -q/(4\pi \epsilon_0 R)$, and therefore express $A_0$ in terms of another constant $q$ with the dimension of charge. The potential is, \[ . While this expression doesn't seem to involve spherical harmonics, in fact they are there in disguise. \tag{10.12} \end{equation}, \begin{equation} Y(y) = e^{\pm i\beta x}, \tag{10.13} \end{equation}, \begin{equation} Y(y) = \left\{ \begin{array}{l} \cos(\beta y) \\ \sin(\beta y) \end{array} \right. The equation is PEspring = 0.5 k x2 where k = spring constant Your solution will be expressed as a Fourier series. Evaluating the potential of Eq. \mu \ddot{r} = -\frac{dU_{\textrm{eff}}}{dr}. We know that $V$ must vanish at $y = \infty$, and this implies that the solution cannot include a factor $e^{\alpha y}$, which grows to infinity. Our result for $\hat{A}_{nm}$ implies that, \begin{equation} 2 A_{nm} = \frac{16 V_0}{\pi^2} \frac{1}{nm} \frac{1}{ \sinh\bigl( \sqrt{n^2+m^2}\, \pi b/a \bigr)}, \tag{10.40} \end{equation}, \begin{equation} V(x,y,z) = \frac{16 V_0}{\pi^2} \sum_{n=1,3,\cdots}^\infty \sum_{m=1,3,\cdots}^\infty \frac{1}{nm} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr) \frac{ \sinh\bigl( \sqrt{n^2+m^2}\, \pi z/a \bigr) }{ \sinh\bigl( \sqrt{n^2+m^2}\, \pi b/a \bigr) } \tag{10.41} \end{equation}. a) What are the factorized solutions for the two-dimensional Laplace equation $\nabla^2 V = 0$ expressed in polar coordinates $s$ and $\phi$? We wish to calculate the electric potential $V$, under the assumption that the conductor is maintained at $V=0$ during the immersion. To represent this constant field at large distances we need a potential that behaves as $V \sim -E z$, or, \begin{equation} V \sim - E\, r \cos\theta, \qquad r \to \infty. The direction of the moment is normal to the plane of the loop . We wish to find $V$ everywhere within the box. Electric Potential Formula. These techniques rest on what was covered in previous chapters. With this restriction on $\mu$, Eq. (10.18) and (10.19), which we write in the hybrid form, \begin{equation} V_{\alpha,\beta}(x,y,z) = \left\{ \begin{array}{l} \cos(\alpha x) \\ \sin(\alpha x) \end{array} \right\} \left\{ \begin{array}{l} e^{i\beta y} \\ e^{-i\beta y} \end{array} \right\} \left\{ \begin{array}{l} e^{\sqrt{\alpha^2+\beta^2}\, z} \\ e^{-\sqrt{\alpha^2+\beta^2}\, z} \end{array} \right\}. for the unique solution to the boundary-value problem, expressed as an infinite sum of products of Bessel and exponential functions. (k - \mu \dot{\phi}^2) r = k\ell \Rightarrow r = \frac{\ell}{1 - (\mu/k) \dot{\phi}^2}. We begin in this chapter with one of the most ubiquitous equations of mathematical physics, Laplace's equation, \begin{equation} \nabla^2 V = 0. U_{\textrm{eff}} = \frac{L_z^2}{2\mu r^2} This is not an initial condition, it's an equilibrium solution. \begin{aligned} It can be shown that for a generic value of $\mu$, the solutions to Eq. You will get the wrong answer! If we begin by considering \( \dot{\phi} \) relatively small, so the denominator remains between 0 and 1, then this expression makes sense: as \( \dot{\phi} \) increases, the equilibrium value of \( r \) is pushed out towards values larger than \( \ell \), where it would be if there was no rotational motion. Connect with a tutor from a university of your choice in minutes. This can be achieved by demanding that $kR$ be a zero of the Bessel function, so that $k = \alpha_{0p}/R$, where, in the notation introduced in Sec.5.3, $\alpha_{0p}$ is the $p^{\rm th}$ zero of the zeroth Bessel function. the energy due to position of a quantity in a field. With $u = \cos\theta$, we are speaking of functions that become infinite at $\theta = \pi$, just like the $Y$ of Eq.(10.73). This is the statement of the superposition principle, and it shall form an integral part of our strategy to find the unique solution to Laplace's equation with suitable boundary conditions. Before the immersion the field was truly constant, but the arrival of the conductor distorts the electric field, because of the induced charge distribution on the surface; the final field is not quite uniform. which we shall insert within Laplace's equation. The electric potential due to a point charge q at a distance of r from that charge is mentioned by: V = q/(4 0 r) In this equation, 0 is the permittivity of free space. U_{\textrm{eff}}(r) = U(r) + \frac{L_z^2}{2\mu r^2}, In this section we suppose that the boundary surfaces are cylinders, and consider solving Laplace's equation using the cylindrical coordinates $(s,\phi,z)$. Find the GPE of an object of mass 10 kg raised 20 m above the ground. JOHN M. COWLEY, in Diffraction Physics (Third Edition), 1995. We shall need the curvilinear coordinates of Chapter 1, the special functions of Chapters 2, 3, 4, 5, and 6, and the expansion in orthogonal functions of Chapters 7, 8, and 9. The list could go on. While Cartesian coordinates are a good choice when Laplace's equation is supplied with boundary conditions on planar surfaces, other coordinates can be better suited to other geometries. 1-519-824-4120 x 52261 \begin{aligned} We will expand on that discussion here as we make an effort to associate the motion characteristics described above with the concepts of kinetic energy, potential energy and total mechanical energy.. Evaluate the potential at $s = \frac{1}{2} R$ and $z = R$. Imagine to be in 2 dimensions and you have to find the potential generated by 4 point-charges of equal charge located at the four corners of a square. Potential energy = (charge of the particle) (electric potential) U = q V U = qV Derivation of the Electric Potential Formula U = refers to the potential energy of the object in unit Joules (J) In general, to have motion with fixed \( r \), the system must be stuck exactly at a single turning point, which means the \( E \) line must be just touching the effective potential curve; this will happen at the minimum possible \( E \). We found that it is a dipole field, with the dipole moment given by = IA, where I is the current and A is the area of the loop. The external work done per unit charge is equal to the change in potential of a point charge. \tag{10.45} \end{equation}. It's easy to sketch what the shape of the potential will look like for various choices of \( L_z \): Why is plotting \( U_{\textrm{eff}} \) useful? (10.32) at $z=0$ yields, \begin{equation} 0 = \sum_{n=1}^\infty \sum_{m=1}^\infty \bigl( A_{nm} + B_{nm} \bigr) \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr), \tag{10.33} \end{equation}, and we find that the expansion coefficients must be related by $B_{nm} = -A_{nm}$. Equation (10.8) states that a function of $x$ only is equal to the sum of a function of $y$ only and a function of $z$ only. Furthermore, the mass of the block of ice is 250 kg. However, that's not really the case. and members of the basis can now be labelled with the integer $n$. \tag{10.1} \end{equation}. \tag{10.3} \end{equation}. (4.44), the differential equation satisfied by the spherical harmonics. The upper hemisphere is maintained at $V = V_0$, while the lower hemisphere is maintained at $V = -V_0$. We could factorize $Y(\theta,\phi)$ further by writing it as $\Theta(\theta) \Phi(\phi)$, but this shall not be necessary. After-lecture aside: what is the equilibrium value of \( r \) here? We cannot expect all solutions to Laplace's equation to be of this simple, factorized form; the vast majority are not. Remember this is completely equivalent to the 1-D problem of finding the motion of \( r \) in such a potential. U_{\textrm{eff}} = -\frac{GM_s m}{r} + \frac{L_z^2}{2\mu r^2}. The factorized solutions become, \begin{equation} V_p(s,z) = J_0(\alpha_{0p} s/R)\, e^{-\alpha_{0p} z/R}, \tag{10.62} \end{equation}. There is a special equation for springs that relates the amount of elastic potential energy to the amount of stretch (or compression) and the spring constant. CUPE 3913 which is exactly the total energy from our original two-dimensional Lagrangian. Changes in membrane potential elicit action potentials and give cells the ability to send messages around the body. Gryph Mail Because of all this freedom, and because $\alpha$ and $\beta$ are arbitrary parameters, we are quite far from having a unique solution to Laplace's equation. Spherical boundaries call for spherical coordinates, and in this section we take on the task of solving Laplace's equation in these coordinates. A potential term is just an ''interaction term'' for scattering experiments so we can add this in in the case of the continuity of eignevalues and [a] solution can be found using the Lippmann-Schwinger equation, just as an example. The idea is to select the blocks that best suit the given problem, and to superpose them so as to satisfy the boundary conditions. Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. The elastic potential energy stored can be calculated using the equation: elastic potential energy = 0.5 spring constant (extension . The more an objects ability to stretch, the greater its elastic potential energy. So both B and D are true. (10.72) becomes, \begin{equation} r^2 \frac{d^2 R}{dr^2} + 2r \frac{dR}{dr} = \ell(\ell+1) R, \tag{10.77} \end{equation}. To achieve this we must require that $\sin(\alpha x) = 0$ at $x = L$, so that $\alpha L$ must be a multiple of $\pi$. For the spring, we find that, \[ Suppose that $V_1$, $V_2$, $V_3$, and so on, are all solutions to Laplace's equation, so that $\nabla^2 V_j = 0$. The factorized solutions to Laplace's equation in spherical coordinates are therefore, \begin{equation} V^m_\ell(r,\theta,\phi) = \left\{ \begin{array}{l} r^\ell \\ r^{-(\ell+1)} \end{array} \right\} Y^m_\ell(\theta,\phi), \tag{10.78} \end{equation}, and they are labelled by the integers $\ell$ and $m$ that enter the specification of the spherical harmonics. If there is no angular momentum, then we only have the second term, which is just the force acting in the radial direction, \( F = -\nabla U \). The only possible way for \( U_{\textrm{eff}} \) to be zero is if both \( r=\ell \) and \( L_z = 0 \) - but this point is precisely the minimum value of \( E \) given those conditions! Physics (Single Science) PSHE and Citizenship . Then show that the expansion coefficients for the constant function $V_0$ are given explicitly by $\hat{A}_{nm} = 16 V_0/(nm \pi^2)$ when $n$ and $m$ are both odd. A generalization to the associated Legendre equation, \begin{equation} (1-u^2) f'' - 2u f' + \biggl[ \lambda(\lambda+1) - \frac{m^2}{1-u^2} \biggr] f = 0, \tag{10.75} \end{equation}. The final solution to our boundary-value problem will be a superposition of these basis solutions. In this case this is the best we can do; there is no simpler expression for the solution, analogous to Eq. It is important to understand that the superposition principle applies to any number of solutions $V_j$; this number could be (and will be) infinite. The third plate is infinite in the $z$-direction only, and it is maintained at $V = V_0$. This is the case here also, as suggested by the fact that the coefficients decrease as $1/n$ with increasing $n$. Work entails the use of a force to shift an object. where $\alpha$ and $\beta$ are arbitrary parameters. Because all plates are infinite in the $z$-direction, nothing changes physically as we move in that direction, and the system is therefore symmetric with respect to translations in the $z$-direction. The boundary conditions and Eq. Again we can go back and forth between the complex exponentials and the trigonometric functions, and the sign in front of $\beta^2$ can be altered by letting $\beta \to i\beta$. But if \( r \) is varying with time, just knowing the minimum and maximum possible values doesn't tell us much about what happens in the middle: Next time: some review, and then we find the equation describing the orbital shape. In particular if I choose the origin of the cartesian coordinates at the center of the square I get (the side . \]. (10.73) becomes <\p> \begin{equation} \frac{1}{\sin\theta} \frac{\partial}{\partial \theta} \biggl( \sin\theta \frac{\partial Y}{\partial \theta} \biggr) + \frac{1}{\sin^2\theta} \frac{\partial^2 V}{\partial \phi^2} = -\ell(\ell+1) Y, \tag{10.76} \end{equation}, which is precisely the differential equation for the spherical harmonics. . \tag{10.9} \end{equation}, The solutions to this ordinary differential equation are, \begin{equation} X(x) = e^{\pm i\alpha x}, \tag{10.10} \end{equation}, \begin{equation} X(x) = \left\{ \begin{array}{l} \cos(\alpha x) \\ \sin(\alpha x) \end{array} \right. = refers to the coefficient of friction = refers to the normal force acting on the object Solved Example on Friction Formula Example 1 Assume a large block of ice is being pulled across a frozen lake. Let's instead turn to the more interesting case of gravitational interaction. (10.63) gives, \begin{equation} V(s,z) = 2 V_0 \sum_{p=1}^\infty \frac{1}{\alpha_{0p} J_1(\alpha_{0p})} J_0(\alpha_{0p} s/R)\, e^{-\alpha_{0p} z/R} \tag{10.68} \end{equation}. We know that \( r \) changes as the particle moves along the line; it decreases to a minimum as \( m_1 \) approaches \( m_2 \), and then starts to increase again. (10.29) satisfies $\nabla^2 V = 0$ and the boundary conditions specified at the beginning of the section. makes the situation even worse. (8.24) informs us that the coefficients are given by, \begin{equation} c_p = \frac{2}{\bigl[ J_1(\alpha_{0p}) \bigr]^2} \int_0^1 V_0 J_0(\alpha_{0p} u)\, u\, du, \tag{10.65} \end{equation}, \begin{equation} c_p = \frac{2V_0}{\bigl[ \alpha_{0p} J_1(\alpha_{0p}) \bigr]^2} \int_0^{\alpha_{0p}} v J_0(v)\, dv \tag{10.66} \end{equation}, by introducing the new integration variable $v := \alpha_{0p} u$. (10.67) within Eq. Enter Displacement. (5.26) reveals that this differential equation is none other than Bessel's equation. At this stage we have obtained that the solution to the boundary-value problem must be built from, \begin{equation} V_k(s,z) = J_0(ks)\, e^{-kz}, \tag{10.61} \end{equation}, The potential is required to go to zero at $s = R$. We write this as, \begin{align} V(x,y,z) &= \sum_{n=1}^\infty \sum_{m=1}^\infty \biggl[ A_{nm} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr) e^{\sqrt{n^2+m^2}\, \pi z/a} \nonumber \\ & \quad \mbox{} + B_{nm} \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr) e^{-\sqrt{n^2+m^2}\, \pi z/a} \biggr], \tag{10.32} \end{align}. The superposition principle follows directly from the fact that Laplace's equation is linear in the potential $V$. Find the GPE of an object of mass 2 kg raised 6 m above the ground. Potential theory. The wall of the pipe is maintained at $V = 0$, and its base is maintained at $V = V_0 (s/R) \sin\phi$. The same condition implies that $B_0 = -A_0 R$, and we now have, \begin{equation} V = A_0 (1 - R/r) - E (r - R^3/r^2) \cos\theta \tag{10.90} \end{equation}, Notice that $V$ depends on an unknown constant $A_0$ that is not determined by the boundary and asymptotic conditions. I've drawn three energy levels on the potential plot. It is the energy generated as a result of an object's location in relation to other items. Collecting results, we find that the factorized solutions to Laplace's equation in Cartesian coordinates are of the form, \begin{equation} V_{\alpha,\beta}(x,y,z) = \left\{ \begin{array}{l} e^{i\alpha x} \\ e^{-i\alpha x} \end{array} \right\} \left\{ \begin{array}{l} e^{i\beta y} \\ e^{-i\beta y} \end{array} \right\} \left\{ \begin{array}{l} e^{\sqrt{\alpha^2+\beta^2}\, z} \\ e^{-\sqrt{\alpha^2+\beta^2}\, z} \end{array} \right\}, \tag{10.18} \end{equation}. Respiration (cellular respiration, that is, not pulmonary respiration or breathing as it is better known) is chemically identical to combustion, but it takes place at a much slower rate. Various forms of energy are studied in physics. Potential difference is also known as voltage. This physics video tutorial provides a basic introduction into kinetic energy and potential energy. (10.37) are given by Eq.(10.39). The potential must also vanish at $x = 0$, and this rules out the presence of a factor $\cos(\alpha x)$. In mathematics and mathematical physics, potential theory is the study of harmonic functions . Find the potential inside the pipe. This idea will be made concrete in the following sections. The charge is given in terms of micro-Coulombs (C): 1.0 C = 1.0 x 10 -6 C. The charge needs to be converted to the correct units before solving the equation: VB = 300 V - 100 V. We recognize the $-E\, r\cos\theta$ contribution to the potential, which gives rise to the constant field at large distances, but we also see a correction proportional to $R^3/r^3$, which comes from the distribution of surface charge on the conductor. This equation is encountered in electrostatics, where $V$ is the electric potential, related to the electric field by $\boldsymbol{E} = -\boldsymbol{\nabla} V$; it is a direct consequence of Gauss's law, $\boldsymbol{\nabla} \cdot \boldsymbol{E} = \rho/\epsilon$, in the absence of a charge density. For example, we need to dig further to answer the simple question: what is the shape of an orbit with a given total energy \( E \)? However, remember that \( \dot{\phi} \) is not arbitrary here! where $c_\ell \propto A^0_\ell$ are the expansion coefficients, and where the factor of $R^{-\ell}$ was inserted for convenience. The Laplacian operator was expressed in these coordinates back in Eq. The solution to this problem will be of the form of Eq. \end{aligned} With $\Phi = e^{\pm im\phi}$, this can be achieved if and only if $2m\pi$ is a multiple of $2\pi$, and this means that $m$ must be an integer, $m = 0, 1, 2, 3, \cdots$. For the gravitational force, the formula is: W = mgh = mgh Where, m is the mass in kilograms g is the acceleration due to gravity h is the height in meters Potential Energy Unit Gravitational potential energy has the same units as kinetic energy: kg m2 / s2 [email protected], College of Engineering & Physical Sciences, College of Social & Applied Human Sciences, Gordon S. Lang School of Business & Economics, Government Relations & Community Engagement. An object which is not raised above the ground will have a height of zero and therefore zero potential energy. (10.45) can be re-expressed as, \begin{equation} m^2 = \frac{s}{S} \frac{d}{ds} \biggl( s \frac{dS}{ds} \biggr) + \frac{s^2}{Z} \frac{d^2 Z}{dz^2}, \tag{10.49} \end{equation}, \begin{equation} -\frac{1}{Z} \frac{d^2 Z}{dz^2} = -\frac{m^2}{s^2} + \frac{1}{sS} \frac{d}{ds} \biggl( s \frac{dS}{ds} \biggr). (10.85) and multiply it by the appropriate factor of $(r/R)^\ell$. Graduate Calendar As a first example of a boundary-value problem formulated in spherical coordinates, we examine a system consisting of two conducting hemispheres of radius $R$ joined together at the equator (see Fig.10.7). This is known as gravitational potential energy. The first is that its solutions are unique once a suitable number of boundary conditions are specified. \]. The answer is that there was no good reason; we made an arbitrary choice, admittedly with an ulterior motive in mind (you'll see). And because the potential should not blow up at the centre of the sphere, we must also kill all $r^{-(\ell+1)}$ terms by setting $B^m_\ell = 0$. Because of the unequal concentrations of ions across a membrane, the membrane has an electrical charge. Each side of the box is maintained at $V=0$, except for the top side, which is maintained at $V=V_0$. The $q/(4\pi \epsilon_0 r)$ term in the potential comes with a correction proportional to $r/R$, and this represents an irrelevant constant. Consider an arbitrary, but finite, charge distribution, ( ), such as that illustrated in Figure (2.1.1). Exercise 10.1: Verify these results for the expansion coefficients $b_n$. The solution is now determined up to the expansion coefficients $A_{nm}$. Proceeding in a similar manner for the function of $y$, we write $g(y) = -\beta^2 = \text{constant}$, or, \begin{equation} \frac{1}{Y} \frac{d^2 Y}{dy^2} = -\beta^2. ), One more note: despite my warning above about Lagrangian substitution, remember that from the perspective of the one-dimensional problem, it's perfectly consistent to treat \( U_{\rm eff} \) as an effective potential energy. The more massive an object is, the greater its gravitational potential energy. (1.75), and we have, \begin{equation} 0 = \nabla^2 V = \frac{1}{s} \frac{\partial}{\partial s} \biggl( s \frac{\partial V}{\partial s} \biggr) + \frac{1}{s^2} \frac{\partial^2 V}{\partial \phi^2} + \frac{\partial^2 V}{\partial z^2} \tag{10.42} \end{equation}, \begin{equation} V = S(s) \Phi(\phi) Z(z), \tag{10.43} \end{equation}, the product of functions of $s$, $\phi$, and $z$. 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