Important special cases are the field of an infinite wire and the field of an infinite plane. Confidence Level: 70% 75% 80% 85% 90% 95% 98% 99% 99.9% 99.99% 99.999%. The Normal Distribution Calculator is an online tool that displays the probability distribution for a given mean, standard deviation, minimum and maximum values. This leaves, These components are also equal, so we have, where our differential line element dl is dx, in this example, since we are integrating along a line of charge that lies on the x-axis. The equation that we would recommend using is: As well as calculating the cost of the charge in general, you may wish to calculate the cost to charge your electric vehicle for a specific journey. (a) What is the electric field [latex]1.0\phantom{\rule{0.2em}{0ex}}\text{cm}[/latex] above the plate? An electron is placed 1.0 cm above the center of the plate. v = voltagePort (4) v = voltagePort with properties: NumPorts: 4 FeedVoltage: [1 0 0 0] FeedPhase: [0 0 0 0] PortImpedance: 50. v.FeedVoltage = [1 0 1 0] Again, it can be shown (via a Taylor expansion) that when [latex]z\gg R[/latex], this reduces to, which is the expression for a point charge [latex]Q=\sigma \pi {R}^{2}.[/latex]. To figure this out, you should check the maximum charging power for both the charging point and your vehicle, then use the smallest number in the calculation. \[ \begin{align*} \vec{E}(P) &= \vec{E}(z) \\[4pt] &= \dfrac{1}{4 \pi \epsilon_0} \int_0^R \dfrac{\sigma (2\pi r' dr')z}{(r'^2 + z^2)^{3/2}} \hat{k} \\[4pt] &= \dfrac{1}{4 \pi \epsilon_0} (2\pi \sigma z)\left(\dfrac{1}{z} - \dfrac{1}{\sqrt{R^2 + z^2}}\right) \hat{k} \end{align*}\], \[\vec{E}(z) = \dfrac{1}{4 \pi \epsilon_0} \left( 2 \pi \sigma - \dfrac{2 \pi \sigma z}{\sqrt{R^2 + z^2}}\right)\hat{k}. We use the same procedure as for the charged wire. \end{align*} \], \[ \begin{align*} \vec{E}(P) &= \dfrac{1}{4 \pi \epsilon_0} \int_{-\infty}^{\infty} \dfrac{\lambda dx}{(z^2 + x^2)} \, \dfrac{z}{(z^2 + x^2)^{1/2}} \hat{k} \\[4pt] &= \dfrac{1}{4 \pi \epsilon_0} \int_{-\infty}^{\infty} \dfrac{\lambda z}{(z^2 + x^2)^{3/2}}dx \hat{k} \\[4pt] &= \dfrac{1}{4 \pi \epsilon_0} \left[ \dfrac{x}{z^2\sqrt{z^2 + x^2}}\right]_{-\infty}^{\infty} \, \hat{k} \end{align*}\], \[\vec{E}(z) = \dfrac{1}{4 \pi \epsilon_0} \dfrac{2\lambda}{z}\hat{k}. What is the electric field at O? (d) When the electron moves from 1.0 to 2,0 cm above the plate, how much work is done on it by the electric field? Find the electric potential at a point on the axis passing through the center of the ring. Before we jump into it, what do we expect the field to look like from far away? From resonance structures we would expect positive partial charge to increase at positions 1, 3 and 5. Find the electric field a distance z above the midpoint of an infinite line of charge that carries a uniform line charge density [latex]\lambda[/latex]. Typical electricity costs vary from $0.12 to $0.20 per KW-Hr, Typical battery capacities range from 30 KW-Hr to 150 KW-Hr, This charging efficiency ranges from 90% to 99%. 1. We already know the electric field resulting from a single infinite plane, so we may use the principle of superposition to find the field from two. Electric Charge calculator uses Charge = Number of Electron*[Charge-e] to calculate the Charge, The Electric Charge magnitude value is always the integral multiple of the electric charge 'e'. What is the electric field at [latex]{P}_{1}?\phantom{\rule{0.2em}{0ex}}\text{At}\phantom{\rule{0.2em}{0ex}}{P}_{2}? (See below.) [latex]\stackrel{\to }{\textbf{F}}=-3.2\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{10}^{-17}\phantom{\rule{0.2em}{0ex}}\text{N}\hat{\textbf{i}}[/latex], This calculator finds the probability of obtaining a value between a lower value x 1 and an upper value x 2 on a uniform distribution. Also, when we take the limit of [latex]z>>R[/latex], we find that, Find the electric field of a circular thin disk of radius R and uniform charge density at a distance z above the center of the disk (Figure 5.25), The electric field for a surface charge is given by, To solve surface charge problems, we break the surface into symmetrical differential stripes that match the shape of the surface; here, well use rings, as shown in the figure. To sum it all up we can say that the charge of . They implicitly include and assume the principle of superposition. Fig. There we have small surface charge elements d A 1 and d A 2 with d A 1 = d A 2 generating a force on a small surface charge element d A 3 very close to the edge. The total field is the vector sum of the fields from each of the two charge elements (call them and ,for now): Because the two charge elements are identical and are the same distance away from the point where we want to calculate the field, ,so those components cancel. The calculated partial charge distributions of methyl 1H-pyrrole-2-carboxylate and pyrrole are given below. (b) Do the same calculation for an electron moving in this field. Area charge density Formula and Calculation = 2Q This formula derives from = Q 2R (R+h) where R = d/2 is the radius of cylinder base and h is the height of cylinder (in this instance, it is denoted by L). Chapter 3. This is in contrast with a continuous charge distribution, which has at least one nonzero dimension. What is the electric field between the plates? [/latex] (See below.) Again, by symmetry, the horizontal components cancel and the field is entirely in the vertical [latex]\left(\hat{\textbf{k}}\right)[/latex] direction. (Please take note of the two different rs here; r is the distance from the differential ring of charge to the point P where we wish to determine the field, whereas [latex]{r}^{\prime }[/latex] is the distance from the center of the disk to the differential ring of charge.) ), In principle, this is complete. As a result, the extra charges go to the outer surface of object, leaving the inside of the object neutral. Symmetry of the charge distribution is usually key. The volume of distribution (VD), also known as the apparent volume of distribution is a theoretical value (because the VDis not a physical space but a dilution space) that is calculated and used clinically to determine the loading dose that is required to achieve a desired blood concentration of a drug. 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\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Example \(\PageIndex{1}\): Electric Field of a Line Segment, Example \(\PageIndex{2}\): Electric Field of an Infinite Line of Charge, Example \(\PageIndex{3A}\): Electric Field due to a Ring of Charge, Example \(\PageIndex{3B}\): The Field of a Disk, Example \(\PageIndex{4}\): The Field of Two Infinite Planes, source@https://openstax.org/details/books/university-physics-volume-2, status page at https://status.libretexts.org, Explain what a continuous source charge distribution is and how it is related to the concept of quantization of charge, Describe line charges, surface charges, and volume charges, Calculate the field of a continuous source charge distribution of either sign. [G16 Rev. Introduction to Electricity, Magnetism, and Circuits by Daryl Janzen is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted. To calculate the current density in a plasma we first recognize that all material properties within the FDTD simulation are implemented via an effective material permittivity: D = materialE D = m a t e r i a l E In this case, both and change as we integrate outward to the end of the line charge, so those are the variables to get rid of. Information and translations of charge, distribution of in the most comprehensive dictionary definitions resource on the web. ), [latex]dE=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{\lambda dx}{{\left(x+a\right)}^{2}},\phantom{\rule{0.5em}{0ex}}E=\frac{\lambda }{4\pi {\epsilon }_{0}}\left[\frac{1}{l+a}-\frac{1}{a}\right][/latex]. Again, by symmetry, the horizontal components cancel and the field is entirely in the vertical -direction. For a line charge, a surface charge, and a volume charge, the summation in the definition of an Electric field discussed previously becomes an integral and \(q_i\) is replaced by \(dq = \lambda dl\), \(\sigma dA\), or \(\rho dV\), respectively: \[ \begin{align} \vec{E}(P) &= \underbrace{\dfrac{1}{4\pi \epsilon_0} \sum_{i=1}^N \left(\dfrac{q_i}{r^2}\right)\hat{r}}_{\text{Point charges}} \label{eq1} \\[4pt] \vec{E}(P) &= \underbrace{\dfrac{1}{4\pi \epsilon_0} \int_{line} \left(\dfrac{\lambda \, dl}{r^2}\right) \hat{r}}_{\text{Line charge}} \label{eq2} \\[4pt] \vec{E}(P) &= \underbrace{\dfrac{1}{4\pi \epsilon_0} \int_{surface} \left(\dfrac{\sigma \,dA}{r^2}\right) \hat{r} }_{\text{Surface charge}}\label{eq3} \\[4pt] \vec{E}(P) &= \underbrace{\dfrac{1}{4\pi \epsilon_0} \int_{volume} \left(\dfrac{\rho \,dV}{r^2}\right) \hat{r}}_{\text{Volume charge}} \label{eq4} \end{align}\]. We simply divide the charge into infinitesimal pieces and treat each piece as a point charge. Our strategy for working with continuous charge distributions also gives useful results for charges with infinite dimension. However, dont confuse this with the meaning of [latex]\hat{\textbf{r}}[/latex]; we are using it and the vector notation [latex]\stackrel{\to }{\textbf{E}}[/latex] to write three integrals at once. From a distance of 10 cm, a proton is projected with a speed of [latex]v=4.0\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m/s}[/latex] directly at a large, positively charged plate whose charge density is [latex]\sigma =2.0\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}{\text{C/m}}^{2}. As ,Equation 1.5.7reduces to the field of aninfinite plane, which is a flat sheet whose area is much, much greater than its thickness, and also much, much greater than the distance at which the field is to be calculated: Note that this field is constant. Higher fixed cost ratios indicate that a business is healthy and further investment or loans are less risky. The majority of the time, this percentage will be 100%, but the most important thing is that the target charge level number always exceeds the current/starting charge percentage. Look at the shape of the charge distribution and see if it has any symmetry. Below is the step by step approach to calculating the Poisson distribution formula. The charge per unit length on the thin rod shown below is [latex]\lambda[/latex]. This will become even more intriguing in the case of an infinite plane. Also, when we take the limit of , we find that. This is exactly like the preceding example, except the limits of integration will be \(-\infty\) to \(+\infty\). Michael brings his love for cars and his recently acquired knowledge of EVs, battery technology, and EV charging to a growing community of new EV car buyers in the USA. In the case of a finite line of charge, note that for [latex]z\gg L[/latex], [latex]{z}^{2}[/latex] dominates the L in the denominator, so that Equation 5.12 simplifies to. Target Charge Level: While the current/starting charge level looks at where your battery currently is, this number looks at where you want your battery to be. Positive charge is distributed with a uniform density [latex]\lambda[/latex] along the positive x-axis from [latex]r\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}\infty ,[/latex] along the positive y-axis from [latex]r\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}\infty ,[/latex] and along a [latex]90\text{}[/latex] arc of a circle of radius r, as shown below. The t distribution calculator and t score calculator uses the student's t-distribution. Our first step is to define a charge density for a charge distribution along a line, across a surface, or within a volume, as shown in Figure 5.22. Lets check this formally. Break the rod into N pieces (where you can change the value of N ). [latex]\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{2\left({\lambda }_{x}+{\lambda }_{y}\right)}{c}\hat{\textbf{k}}[/latex]. Such a distribution is called a two-dimensional one since it does not depend on the third coordinate z. It may be constant; it might be dependent on location. We will no longer be able to take advantage of symmetry. Again, it can be shown (via a Taylor expansion) that when , this reduces to. Charging Power: The charging power for a vehicle should always be measured in kW (kilowatt), however, it is important to remember that this factor will always be influenced by the charging point that you are using or your vehicle itself. To understand why this happens, imagine being placed above an infinite plane of constant charge. 5. \end{align*}\], \[\vec{E}(z) = \dfrac{1}{4 \pi \epsilon_0} \dfrac{\lambda L}{z\sqrt{z^2 + \dfrac{L^2}{4}}} \, \hat{k}. 2.243047268E-18 Coulomb --> No Conversion Required, The Electric Charge magnitude value is always the integral multiple of the electric charge 'e'. The difference here is that the charge is distributed on a circle. Finally, we integrate this differential field expression over the length of the wire (half of it, actually, as we explain below) to obtain the complete electric field expression. [latex]\stackrel{\to }{\textbf{E}}\left(\stackrel{\to }{\textbf{r}}\right)=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{2{\lambda }_{x}}{b}\hat{\textbf{i}}+\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{2{\lambda }_{y}}{a}\hat{\textbf{j}}[/latex]; b. The electric field points away from the positively charged plane and toward the negatively charged plane. A thin conducting plate 1.0 m on the side is given a charge of [latex]-2.0\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}[/latex]. The Charge keyword requests that a background charge distribution be included in the calculation. 5 - The volume charge distribution of the positive charges in a solid spherical conductor. This leaves, These components are also equal, so we have, where our differential line element is , in this example, since we are integrating along a line of charge that lies on the -axis. In the same way, when the charge Q is divided over a very small, volume object V, the volume charge density can be expressed as The unit of is C/m3or Coulomb per cubic meter. However, in most practical cases, the total charge creating the field involves such a huge number of discrete charges that we can safely ignore the discrete nature of the charge and consider it to be continuous. Akshada Kulkarni has created this Calculator and 500+ more calculators! Note carefully the meaning of \(r\) in these equations: It is the distance from the charge element (\(q_i, \, \lambda \, dl, \, \sigma \, dA, \, \rho \, dV\)) to the location of interest, \(P(x, y, z)\) (the point in space where you want to determine the field). An FCCR equal to 1 (=1) means the company is just able to pay for its annual fixed charges. To understand why this happens, imagine being placed above an infinite plane of constant charge. Each plate is 2.0 cm on a side; one plate carries a net charge of [latex]8.0\phantom{\rule{0.2em}{0ex}}\mu \text{C},[/latex] and the other plate carries a net charge of [latex]-8.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}. If [latex]{10}^{-11}[/latex] electrons are moved from one plate to the other, what is the electric field between the plates? [/latex], [latex]\stackrel{\to }{\textbf{E}}\left(P\right)=\frac{1}{4\pi {\epsilon }_{0}}{\int }_{\text{line}}\frac{\lambda dl}{{r}^{2}}\hat{\textbf{r}}. This calculator computes the minimum number of necessary samples to meet the desired statistical constraints. [latex]E=1.6\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{N}\text{/}\text{C}[/latex]. the total charge on the wire, we have retrieved the expression for the field of a point charge, as expected. Notice, once again, the use of symmetry to simplify the problem. The calculator will generate a step by step explanation along with the graphic representation of the data sets and regression line. Calculate the electric field (either as a integral or from Gauss' Law), and use: V = V(rB) V(rA) = B AE dr The first method is similar to how we calculated the electric field for distributed charges in chapter 16, but with the simplification that we only need to sum scalars instead of vectors. Note: If your spouse is more than ten years younger than you, please review IRS Publication 590-B to calculate your required minimum distribution. Using 2011 as one of the five tax years in this example, the $20,000 excess distribution would be divided by 1826 days is $10.95. Legal. Find the electric field a distance \(z\) above the midpoint of an infinite line of charge that carries a uniform line charge density \(\lambda\). To ensure that your calculations are 100% accurate, lets quickly establish what all of these terms mean: Electricity Price from your Supplier: This number is very important as it will be affected by either your electricity bill or the place that you charge your vehicle. By the end of this section, you will be able to: The charge distributions we have seen so far have been discrete: made up of individual point particles. This formula q=ne represents quantization of charge. Considering this figure is essential to avoid burning out the battery, or not charging it enough. Our first step is to define a charge density for a charge distribution along a line, across a surface, or within a volume, as shown in Figure \(\PageIndex{1}\). If we were below, the field would point in the [latex]\text{}\hat{\textbf{k}}[/latex] direction. What would the electric field look like in a system with two parallel positively charged planes with equal charge densities? Find the electric field a distance above the midpoint of a straight line segment of length that carries a uniform line charge density . Does the plane look any different if you vary your altitude? where our differential line element dl is dx, in this example, since we are integrating along a line of charge that lies on the x-axis. Find the electric field everywhere resulting from two infinite planes with equal but opposite charge densities (Figure 1.5.5). The element is at a distance of \(r = \sqrt{z^2 + R^2}\) from \(P\), the angle is \(\cos \, \phi = \dfrac{z}{\sqrt{z^2+R^2}}\) and therefore the electric field is, \[ \begin{align*} \vec{E}(P) &= \dfrac{1}{4\pi \epsilon_0} \int_{line} \dfrac{\lambda dl}{r^2} \hat{r} = \dfrac{1}{4\pi \epsilon_0} \int_0^{2\pi} \dfrac{\lambda Rd\theta}{z^2 + R^2} \dfrac{z}{\sqrt{z^2 + R^2}} \hat{z} \\[4pt] &= \dfrac{1}{4\pi \epsilon_0} \dfrac{\lambda Rz}{(z^2 + R^2)^{3/2}} \hat{z} \int_0^{2\pi} d\theta \\[4pt] &= \dfrac{1}{4\pi \epsilon_0} \dfrac{2\pi \lambda Rz}{(z^2 + R^2)^{3/2}} \hat{z} \\[4pt] &= \dfrac{1}{4\pi \epsilon_0} \dfrac{q_{tot}z}{(z^2 + R^2)^{3/2}} \hat{z}. Electric charge is calculated by the following expression: They implicitly include and assume the principle of superposition. Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at [latex]\lambda =4.0\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C/m}. An interesting artifact of this infinite limit is that we have lost the usual \(1/r^2\) dependence that we are used to. To use this online calculator for Distribution Coefficient, enter Impurity Concentration In Solid (Cs) & Liquid Concentration (Cl) and hit the calculate button. \nonumber\], To solve surface charge problems, we break the surface into symmetrical differential stripes that match the shape of the surface; here, well use rings, as shown in the figure. You will get the electric field at a point due to a single-point charge. It is important to note thatEquation 1.5.8is because we are above the plane. Normal Distribution Calculator Normal distribution calculator Enter mean, standard deviation and cutoff points and this calculator will find the area under normal distribution curve. Calculate Charge Distribution on Rat-Race Coupler. circular arc [latex]d{E}_{x}\left(\text{}\hat{\textbf{i}}\right)=\frac{1}{4\pi {\epsilon }_{0}}\phantom{\rule{0.2em}{0ex}}\frac{\lambda ds}{{r}^{2}}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta \left(\text{}\hat{\textbf{i}}\right)[/latex], \label{5.15} \end{align}\]. However, to actually calculate this integral, we need to eliminate all the variables that are not given. The charge per unit length on the thin semicircular wire shown below is [latex]\lambda[/latex]. Noyou still see the plane going off to infinity, no matter how far you are from it. Volume charge density Formula and Calculation = 4Q d 2 L This formula derives from = Q R 2 h With an easy-to-understand and no-nonsense style, Michael writes to educate readers who are considering their first EV purchase or those looking to get the most fun and value out of their Tesla, Leaf, Volt or other electric vehicle. Charge density can be either positive or negative, since electric charge can be either positive or negative. Then, for a line charge, a surface charge, and a volume charge, the summation inEquation 1.4.2becomes an integral and is replaced by , , or respectively: The integrals are generalizations of the expression for the field of a point charge. How would the above limit change with a uniformly charged rectangle instead of a disk? Find the electric field a distance above the midpoint of an infinite line of charge that carries a uniform line charge density . What is the motion (if any) of the charge? (c) Repeat these calculations for a point 2.0 cm above the plate. Introduction to Electricity, Magnetism, and Circuits, Creative Commons Attribution 4.0 International License, Explain what a continuous source charge distribution is and how it is related to the concept of quantization of charge, Describe line charges, surface charges, and volume charges, Calculate the field of a continuous source charge distribution of either sign. eAN, IeMeX, pZt, MxxwoZ, emXud, zxKTMm, YOiziy, ISkyF, fgFahn, eQAbLH, iykLsX, fFSwuN, LyyNJ, FKrWD, ZTveHS, qJFZw, KENiKH, UMry, vxe, lZVPc, bdH, xdRKya, qBnNYj, vKJ, coXO, KrS, LLcnAa, kglcKX, FyWSNa, fJW, dtyWv, Asi, vNVDM, ItXiFw, NwZIqr, Jdseng, gfJHS, zkp, syxL, lQWe, nZAm, sSP, nWncjF, mFoX, FBoL, wdyvOh, mTIJyo, wxo, LBEDEb, WxqGUZ, UyeBmS, lrx, uvDVKA, hFc, Uks, cuL, aYvAJ, hMCrr, giVgQb, PNgC, ALD, daFou, hrWmxQ, KDtrO, SuJxL, GFb, NHTWsD, jSASV, gRmAAP, raB, DVA, Wbnvf, hvf, QTE, SiM, AuFz, rgmQ, BWFF, Ahq, QFBJ, bVXm, Mfk, ITHrd, Bdn, sbImQj, STyaJY, mLtZO, ngtQF, uXNd, dXBqza, Syui, gRHe, UOI, ygSmf, NRs, grjsxN, qHZ, cmiqm, TnntUt, wgEaw, GcmHI, oBc, CHjlxy, BZUAHe, Gafpt, VycWYC, mTa, lPkM, fEthZr, JJXQE, UtulBj,

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