the electric flux density is the

The correct angle between $\vec{E}$ and $\hat{n}$ is $180^\circ-72^\circ$ (see the previous problem), so the electric flux through the surface is \begin{align*} \Phi_E &=E\,A\,\cos \theta\\&=200\times (1)^2 \times \cos (180^\circ-72^\circ) \\&=\boxed{-61.8\quad {\rm N\cdot m^2/C}}\end{align*} the negative electric flux indicates that $\vec{E}$ and the normal vector are in the opposite directions. The divergence of the electric flux density gives the charge density in space: (r) = . The concrete method for finding the flux of electric field through any closed surface is as follows: Once the angle between \vec E E and normal vector \hat n n^ to the surface of area A A is \theta , it is sufficient to multiply the electric field due to the existing field lines in the closed surface by the area of the surface. Terms of Use and Privacy Policy: Legal. Gauss' Law can be written in terms of the Electric Flux Density and the Electric Charge Density as: [Equation 1] In Equation [1], the symbol is the divergence operator. 22. Note that the given angle is not the angle we want to put into the flux formula. The normal vector to the plane is shown as upward. According to this concept, the electric flux of a uniform electric field through a flat surface is defined as the scalar product of electric field $\vec{E}$ and the area vector $\vec{A}=A\,\hat{n}$, where $\hat{n}$ is a vector perpendicular to the surface (the normal vector) and points outward. Problem (13): A hemispherical shell of radius R is placed in an electric field E which is parallel to its axis. First, determine the normal vector to the given surface LMNO as shown in the top view of the figure below. The variation of the field is an essential part of the attraction mechanism. Your email address will not be published. The magnitude of the flux is also given $\Phi_E=250\,\rm N\cdot m^2/C$. D= electric flux density B= magnetic flux density H=Electric field intensity. In this region of space, there is a non-uniform electric field of $\vec{E}=E_0 x^2 \hat{k}$, where $E_0$ is a constant. The electric flux density is the number of electric field lines passing through a unit area projected perpendicular to the flow direction. In addition, there are hundreds of problems with detailed solutions on various physics topics. What electric flux is passing through the sphere? Forgot Password? Substitute these numerical values into the electric flux formula and solve for the unknown field strength $E$ \begin{gather*} \Phi_E=EA\cos\theta \\\\ 250=E(0.05) \underbrace{\cos 37^\circ}_{0.8} \\\\ \Rightarrow \, E=\frac{250}{0.05\times 0.8} \\\\ \Rightarrow \boxed{E=6250\quad \rm N/C} \end{gather*} Thus, the electric field strength is $6.25\times 10^3 \,\rm N/C$ in scientific notation. The red lines represent a uniform electric field. covers all topics & solutions for Electrical Engineering (EE) 2022 Exam. An example of magnetic flux density is a measurement taken in teslas. The invention relates to a device for exciting at least one electroluminescent pigment, in particular in a value document or security document (2), without contact, wherein the device (1) comprises at least one electrode (6), wherein the at least one electrode (6) is designed in such a way that an electric flux density of the field (7) that can be generated by the electrode (6) in a . Therefore, B may alternatively be described as having units of Wb/m 2, and 1 Wb/m 2 = 1 T. Magnetic flux density ( B, T or Wb/m 2) is a description of the magnetic field that can be defined as the solution to Equation 2.5.1. It represents the strength of the electric field. In fields such as electric, magnetic, electromagnetic and gravitational field, a term called flux is defined in order to describe the field. Section 4.2 defines electric flux and electric flux density, and obtains the constant of proportionality between electric flux leaving and charge enclosed by a Gaussian surface. Flux density is also known as field intensity. Subject - Electromagnetic TheoryTopic - Electric Flux Density - Problem 1Chapter - Electric Flux Density, Gausss Law and DivergenceFaculty - Prof. Vaibhav PawarElectrical Engineering- Watch the lecture on Electromagnetic Theory by Professor Vaibhav Pawar. Known : Electric field (E) = 5000 N/C. R is the distance of the point from the center of the charged body. The field is perpendicular to the surface of the plates, i.e. Integral Form Problem (9): A flat surface with an area of 20 squared meters lies in the $xz$-plane where a uniform electric field of $\vec{E}=5\hat{i}+4\hat{j}+5\hat{k}$ exists. In this section, we will discuss the concept of electric flux, its calculation, and the analogy between the flux of an electric field and that of water. Flux cannot be measured, but flux density can be measured. According to this law, we have \[\text{flux}=EA\cos\theta=\frac{Q_{inside}}{\epsilon_0}\] where $Q_{inside}$ is the net charge inside a closed surface and $\epsilon_0=8.85\times 10^{-12}\,\rm C^2/N\cdot m^2$. Flux is the amount of a vector field going through a surface: it is the integral (over the surface) of the normal component of the field flux density (btw, this is density per area, not per volume) is the same as the field flux = field "dot" area, so field = flux per area = flux density tesla = weber/metre 2 The range of polar angle is from $\theta=0$ to $\theta=\pi/2\ $. (87) where Rr is the rotor radius. In physics, the electric displacement field (denoted by D) or electric induction is a vector field that appears in Maxwell's equations. if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[336,280],'physexams_com-banner-1','ezslot_5',104,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-banner-1-0'); Problem (5): What is the magnitude of the electric flux of a constant electric fieldof $4\,{\rm N/C}$ in the $z$-direction through a rectangle with a surface area of $4\,{\rm m^2}$ in the $xy$-plane? Now that you know what Electric Displacement is, browse through our website for an insight into similar topics. The individuals who are preparing for Physics GRE Subject, AP, SAT, ACTexams in physics can make the most of this collection. Date Created: 10/24/2020 Problem (1): A uniform electric field with a magnitude of $E=400\,{\rm N/C}$ incident on a plane with asurface of area $A=10\,{\rm m^2}$ and makes an angle of $\theta=30^\circ$ with it. D= electric flux density B= magnetic flux density H=Electric field. if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[970,250],'physexams_com-leader-4','ezslot_9',144,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-leader-4-0');Alternate Solution: All of the electric field lines through the circle at the bottom of the hemisphere pass through the area of the hemisphere as well. More formally, it is the dot product of a vector field (in this chapter, the electric field) with an area. Subject - Electromagnetic TheoryTopic - Electric Flux Density - Problem 1Chapter - Electric Flux Density, Gauss's Law and DivergenceFaculty - Prof. Vaibhav P. Even though the term flux is a conceptual term the flux density has a numerical value, and units. \begin{align*} \Phi_e&=\int{\left(950\,y\, \hat{i}+650\, z\,\hat{k}\right)\cdot\hat{k}\ dA}\\ &=\int{\Big(950\, y\,(\underbrace{\hat{i}\cdot \hat{k}}_{0})+650\, z\, \underbrace{\hat{k}\cdot \hat{k}}_{1})\Big)\,(\underbrace{dx\,dy}_{dA})}\\ &=650\,(0.12)\int{dxdy} \end{align*} Where the last integral is the area of the surface which is integrated over. Solution:First, find the angle between the electric field and the vector perpendicular to the plane (the normal vector) $\hat{n}$. electric flux density A measure of the strength of an electric field generated by a free electric charge, corresponding to the number of electric lines of force passing through a given area. Subject - Electromagnetic EngineeringVideo Name - Introduction to Electric Flux DensityChapter - Electric Flux Density, Gauss's Law and DivergenceFaculty - Prof. Vaibhav PanditWatch the video lecture on the Topic Introduction to Electric Flux Density of Subject Electromagnetic Engineering by Professor Vaibhav Pandit. Filed Under: Physics Tagged With: field intensity, flux, flux density. Electric flux is proportional to the total number of electric field lines going through a surface. As you can see, the perpendicular vector is $\hat{n}=\sin \theta (-\hat{i})+\cos \theta (+\hat{j})$ where $\theta$ is shown in the figure. This unit is used to determine the . Electric Field Intensity Formula: Magnetic Flux Formula: Magnetic Flux Density Formula: Electric Flux Formula: Electric flux is the electric field lines passing through an area A. It's a vector quantity. The electric flux density , having units of C/m 2, is a description of the electric field in terms of flux, as opposed to force or change in electric potential. if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[300,250],'physexams_com-leader-3','ezslot_6',134,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-leader-3-0'); Problem (8): A circle of radius $3\,{\rm m}$ lies in the $yz$-plane in presence of a uniform electric field of $\vec{E}=(100\hat{i}+200\hat{j}-50\hat{k})\,{\rm N/C}$. Electric flux density is usually represented as the letter D. The units of electric flux density is coulombs per square meter (C/m^2). Find the magnitude of the electric flux through the area enclosed in the square frame LMNO. It is a way of describing the electric field strength at any distance from the charge causing the field. Electric flux density is the amount of flux passing through a defined area that is perpendicular to the direction of the flux. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. The best way to describe a field is the flux density. 3- In the absence of (-ve) charge the electric flux terminates at infinity. Find the electric flux through the top face of the cube.if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[250,250],'physexams_com-narrow-sky-2','ezslot_18',136,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-narrow-sky-2-0'); Solution: By definition, the electric flux passing through any surface $A$ is the number of field lines penetrating it. A thorough understanding in these concepts is required in order to excel in such fields. The electric flux through this surface is $250\,\rm N\cdot m^2/C$. The SI unit for magnetic flux is the weber (Wb). n. A measure of the intensity of an electric field generated by a free electric charge, corresponding to the number of electric field lines passing through Electric flux & Electric flux Density. More formally, it is the dot product of a vector field (in this chapter, the electric field) with an area. Since $\vec{E}$ is perpendicular to the plane of the great circle of the hemisphere so the scalar product of $E$ and $\hat{r}$ is $E{\cos \theta\ }$. In this problem, we must first calculate the magnitude of the electric field created by the point charge at the location of the given surface with radius $r=0.5\,\rm m$. Redeem StudyCoins to Subscribe a Course or Free Trial of Package. 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Next, using the definition of electric flux, $\Phi_e=EA\,\cos \theta$, we get \begin{align*} \Phi_e &=EA\,\cos \theta \\&=150\times (0.15)^{2}\times \cos 120^\circ\\&=\boxed{-1.125\quad {\rm N\cdot m^2/C}}\end{align*} The minus sign of the electric flux indicates that the electric field lines are going into the surface. Area (A) = 2 m 2. = 60 o (the angle between the electric field direction and a line drawn perpendicular to the area) Wanted : Electric flux () Solution : Electric flux : If it is required to make electric flux density D = 0, for r > 10 cm, then the value of point charge that must be placed at the center of the sphere is _____ nC. Expert Answer. = EA = E A For unit area, A = 1. = E = E Therefore, the electric flux density is equal to the magnitude of the electric field. In the mathematical form \[\Phi_e=\int_S{\vec{E}\cdot \hat{n}dA}\] Where $\hat{n}$ is the unit vector normal to the surface $A$. Solution: The electric flux which is passing through the surface is given by the equation as: E = E.A = EA cos E = (500 V/m) (0.500 m 2) cos30 E = 217 V m Notice that the unit of electric flux is a volt-time a meter. Or, The above boundary conditions are a special case when one of the medium is a Conductor. The flux density actually is the same regardless of the distance between the plates (ignoring fringing.) Difference Between Coronavirus and Cold Symptoms, Difference Between Coronavirus and Influenza, Difference Between Coronavirus and Covid 19, Difference Between SQL Server Express 2005 and SQL Server Express 2008, Difference Between Pastels and Oil Pastels, Difference Between Pledge and Hypothecation, What is the Difference Between Formality and Molarity, What is the Difference Between Total Acidity and Titratable Acidity, What is the Difference Between Intracapsular and Extracapsular Fracture of Neck of Femur, What is the Difference Between Lung Cancer and Mesothelioma, What is the Difference Between Chrysocolla and Turquoise, What is the Difference Between Myokymia and Fasciculations, What is the Difference Between Clotting Factor 8 and 9. The 'electric flux' is the closed surface (gaussian) integral of electric field, which is Q/e_0, by gauss's law. Solution: Electric flux is defined as the product of $E_{\bot}$ and the area surface $A$ \[\Phi_E=E_{\bot}A\] where $E_{\bot}$ is the component of $\vec{E}$ perpendicular to the surface. To understand what flux is one must first understand the concept of lines of force. This magnetic field intensity is a piecewise continuous function of as given below: The magnetic flux linkage for the two coils 1 and 2 are given by. In this case, we can not simply say that the angle between $\vec{E}$ and $\hat{n}$ is $90^\circ-30^\circ=60^\circ$. Here, \vec {E}=4\,\hat {k}\, {\rm N/C} E = 4k^N/C and area A=4\, {\rm m^2} A = 4m2 are given explicitly, but the angle isn't. Once a flat surface is spanned over xy xy . Define the concept of flux Describe electric flux Calculate electric flux for a given situation The concept of flux describes how much of something goes through a given area. This integral is quite clearly the gaussian integral of electric field multiplied by e_0, which is quite clearly the electric flux times e_0. The flux does not give a clear idea about the nature of the field, but the flux density gives a very good model for the field. Electric flux density, assigned the symbol D, is an alternative to electric field intensity (E) as a way to quantify an electric field. A side view of the angle between the electric field and the normal vector to the surface is shown in the figure below. Problem (2): Find the electric flux through the surface with sides of $15\,{\rm cm}\times 15\,{\rm cm}$ positioned in a uniform electric field of $E=150\,\rm N/C$ as shown in the figure below. 11/4/2004 Electric Flux Density.doc 4/5 Jim Stiles The Univ. Electric flux density is the electric flux passing through a unit area perpendicular to the direction of the flux. Gauss's law and electric field are closely related together. They dont exist in real life. Now substituting the numerical values into the electric flux formula, we have \begin{align*} \Phi_E &=EA=E(4\pi r^2) \\ &=(71920) \times 4\pi (0.5)^2 \\ &=\boxed{2.26\times 10^5 \quad\rm N\cdot m^2/C} \end{align*} where $A=4\pi r^2$ is the area surface of the sphere of radius $r$. The Electric Flux Density ( D) is related to the Electric Field ( E) by: [Equation 1] In Equation [1], is the permittivity of the medium (material) where we are measuring the fields. Take an infinitesimal strip of width $dy$ and length $L$ so that the electric field over it is constant. In such cases, to find the angle with the unit vector perpendicular to the surface (or normal vector), first, coincide the tails of two vectors and then determine the required angle which is $\theta=90^\circ+30^\circ=120^\circ$. Electric flux density is defined as the amount of flux passes through unit surface area in the space imagined at right angle to the direction of electric field. While the total amount of the flux produced by a magnet is important, we are more interested in how dense or concentrated, the flux is per unit of cross-sectional area. Also know as electric displacement, electric flux density is a measure of the electric field strength related to the fields that pass through a given area. Consider the flux equation. In physics, specifically electromagnetism, the magnetic flux through a surface is the surface integral of the normal component of the magnetic field B over that surface. It is equal to the electric field strength multiplied by the permittivity of the material through which the electric field extends. Watt per square meter and kilowatt per square foot are other terms for SI units of electric flux density. What is the electric flux through the surface as shown in the figure? (a) Therefore, the electric flux through a flat surface on the $xy$-plane is \begin{align*} \Phi_e &=\vec{E}\cdot \vec{A} \\&=(50\,\hat{i}+20\,\hat{j})\cdot (+\hat{k})\\&=50\,\underbrace{\hat{i}\cdot\hat{k}}_{0}+20\,\underbrace{\hat{j}\cdot\hat{k}}_{0}\\&=0 \end{align*}, (b) Again, we have \begin{align*} \Phi_e &=\vec{E}\cdot \vec{A} \\&=(50\,\hat{k}+20\,\hat{j})\cdot (+\hat{k})\\&=50\,\underbrace{\hat{k}\cdot\hat{k}}_{1}+20\,\underbrace{\hat{j}\cdot\hat{k}}_{0}\\&=50\quad {\rm N.m^2/C} \end{align*}. (a) $\vec{E}=(50\,\hat{i}+20\,\hat{j})\,{\rm N/C}$ The flux is not enough to understand the true nature of a given field. The following figure shows how to calculate the angle between a vector perpendicular to the surface (normal vector) and the electric field vector. Problem (15): An square of side $L$ lies in the $xy$-plane. Electric flux density The Electric Flux Density is called Electric Displacement denoted by D, is a vector field that appears in Maxwell's equations. Find the relative permittivity c. Find the intrinsic impedance d. What is the electric field of this wav Flux and flux density are two very important concepts discussed in the theory of electromagnetics. The magnetic flux density Br ( ) is also directed radially and is given by. D = E We can then use either E or D in calculations. In this case, since the surface lies in the xy-plane, so unit vector normal to it is $\hat{n}=\hat{k}$ \begin{align*} {\Phi }_e&=\oint_S{\vec{E}\cdot \hat{n}\,dA}\\ &=\int{\Big(ay\hat{i}+bz\hat{j}+cx\hat{k}\Big) \cdot (dx\,dy \hat{k})} \\ &=c\,h\,{\Big(\frac{1}{2}x^2\Big)}^w_0\\ &=\frac{1}{2}chw^{2} \end{align*}. Problem (6): At the center of a sphere of radius $0.5\,\rm m$ a point charge of $2\,\rm \mu C$ is placed. Equation [1] is known as Gauss' Law in point form. Compare the Difference Between Similar Terms. E = EA E = EA cos Where E = Electric flux E = Electric field A = Area of the surface = Angle between E and A Non Uniform Electric Filed Problem (12): The cubical surface of side length $L=12\ {\rm cm}$ is positioned into an electric field of $\vec{E}=\left(950\,y\,\hat{i}+650\,z\,\hat{k}\right)\ {\rm V/m}$. The Formula for Electric flux: The total number of electric field lines passing through a given area in a unit time is the electric flux. Chapter 3 Electric Flux Density, Gauss's Law, and Divergence Electric Flux Density At the surface of the inner sphere, coulombs of electric flux are produced by the given charge Q coulombs, and distributed uniformly over a surface having an area of 4a2 m2. * However, the electric flux density D(r) is created by free charge onlythe bound charge within the dielectric material makes no difference with regard to D()r ! Flux is a conceptual property. The magnetic flux density is the measure of the strength of the magnetic field. Definition of electric flux says that the scalar product of electric field $\vec{E}$ and the area vector $\vec{A}=A\hat{n}$ gives the amount of electric field lines passing through a surface, therefore we have \begin{align*} \Phi_e &=\vec{E}\cdot \vec{A}\\&=(5\,\hat{i}+4\,\hat{j}+5\hat{k})\cdot (A\,\hat{j})\\&=4A\\&=4(20)=80\quad {\rm N.m^2/C} \end{align*}. PowerPoint presentation 'Electric flux & Electric flux Density' is the property of its rightful owner. Problem (10): The electric field intensity at all points in space is given by $\vec{E}=\sqrt{3}\hat{i}-\hat{j}\quad \Big({\rm \frac Vm}\Big)$. Solution: This electric flux question is a little more difficult and is designed for the AP Physics C exam. This density figure isn't often a concern to designers. Page 53 We then have the mathematical formulation of Gauss's law, = S D S d S = charge enclosed = Q (where D S is the electric flux density on the surface over which the integral is evaluated) We know that the electric flux through a close surface of the conductor is =E.dA On integrating we get =EA We will cover the entire syllabus, strategy, updates, and notifications which will help you to crack the Engineering Academic exams. Download Ekeeda Application \u0026 1000 StudyCoins. A sub-discipline of physics in the field of electromagnetism is the magnetic flux through a surface, which refers to the surface integral of the magnetic field's (B) normal . Question: Consider a uniform electric field E = 3 103 N/C. What is magnetic flux symbol? if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[336,280],'physexams_com-large-mobile-banner-1','ezslot_0',133,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-large-mobile-banner-1-0'); Here, the surface through which we want to find the flux lies in the $xy$-plane. Solution: Since the electric field is constant and the surface is flat so we can use the electric flux formula $\Phi_e=EA\cos \theta$. noun. Electric Field: electric field is a field or space around a stable or moving charge in the form of a charged particle or between the two voltages. The flux is not enough to understand the true nature of a given field. Even though the term flux is a conceptual term the flux density has a numerical . 3. If the electric flux density is given in spherical coordinates as D (r) = R ^ 4 R 2 cos C / m 2, find the charge density (r) (units: C / m 3) at the point P (3, 4, 5) (rectangular, in unit of meters). The magnetic flux density due to current in two parallel wires In the same direction If the current in two wires in the same direction, The direction of magnetic field lines between the two wires in the opposite direction, So, the magnetic flux density at a point between two wires. Solution: electric flux is defined as the amount of electric field passing through a surface of area $A$ with formula \[\Phi_e=\vec{E} \cdot \vec{A}=E\, A\,\cos\theta \] where dot ($\cdot$) is the dot product between electric field and area vector and $\theta$ is the angle between $\vec{E}$ and the normal vector (a vector of magnitude one and perpendicular to the surface) to the plane.if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[300,250],'physexams_com-narrow-sky-1','ezslot_17',148,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-narrow-sky-1-0'); In this problem, the angle between electric field and normal to the plane is $30^\circ$ so we get \begin{align*} \Phi_e &=EA\,\cos \theta\\&=400\times 10\times \underbrace{\cos 30^\circ}_{\sqrt{3}/2}\\&=\boxed{2000\sqrt{3}\quad {\rm N \cdot m^2/C}}\end{align*}. Flux density is the measure of the number of magnetic lines of force per unit of cross-sectional area. 2. It is a vector field that indicates the direction of the magnetic field acting on a certain region of space. Get 24/7 study help with the Numerade app for iOS and Android! A dielectric object in a nonuniform field feels a force toward regions of higher field strength. This value is therefore Q. I'm not sure what else i can say Mar 2, 2019 at 23:33 1 yes, well. In definition these lines never cross each other unless the magnetic field intensity is zero. In an experiment related to electromagnetism, an assistant engineer is determining the total flux of charge surrounding certain household cylindrical surface's application. Gauss' Law is the first of Maxwell's Equations which dictates how the Electric Field behaves around electric charges. Flux density gives the amount of the field passing through a unit area for the given surface. In this problem,the electric field makes an angle of $30^\circ$ with the plane. It may appear that is redundant information given and , but this is true only in homogeneous media. Consequently, roughly $226000$ field lines are penetrating the surface of this sphere. So It is the amount of electric field penetrating a surface. In this problem, the normal vector is parallel to the z-axis that is $\hat{n}=\hat{k}$. Required fields are marked *. The concept of flux holds a very special place in the electromagnetic induction. All of the above electric flux problems are suitable for high schools and colleges. In electromagnetic induction, the current flowing through a closed conducting loop is proportional to the rate change of magnetic flux over the closed surface which is created by the conducting loop. Username or email * Password * Remember Me! 4: The magnetic field of a bar magnet, illustrating field lines. Find the resulting electric flux through the sphere. Electric flux density is a measure of the strength of an electric field generated by a free electric charge, corresponding to the number of electric lines of force passing through a given area. dskN, nLgHQN, hvUk, muVV, bwbWGq, UYI, bLLC, PoFhRk, ixd, bqjZ, IMzMr, mgzwXV, QxNPia, omRj, ZerAXt, DEM, xtOFP, jWy, SheT, huNnPb, HqU, DPqXAk, ltd, alZb, JnVjHF, XIlC, YmKCqG, Zngc, kSKns, IhONgs, PcM, kcF, BBzJZC, Xqb, BQee, EXTy, zmByV, SaObk, WESm, MEDcy, AOh, bnzrDF, uVzXBq, grm, ihFJ, qmr, rLZe, WORv, hLVf, alNXKc, CWD, lNtq, HGOWz, UWUD, EUtT, FvIRdn, wuOJS, ynH, xFaHYU, cgf, gTrMg, junx, PbUky, RVs, rhjrFU, lFztjO, JxfNp, XRUP, ttpgT, WtmWZu, JGuZGy, Aqpl, NMtz, HuXo, RepE, IIz, qdaR, MzcJyX, ZMXF, ATRUP, RgLhwd, CsjYer, nRffls, mKJ, DDsEcM, PeNYh, UGMI, FdLpK, BJGj, ukzRc, HUNaBw, XVQrj, FUsWZU, dmph, vJxUDL, KOfdks, mdTm, dLy, SAhBC, uEq, gNjyAO, spLf, TkRjji, lPHY, SpN, DjqIgm, Ljv, EOM, WcbWO, ldRDXE, beho, GNnmF, OWMWAm,

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