$$ We calculate an electrical field of an infinite sheet. then, we "set the zero of potential at infinity." You know that if you have a point charge with charge $Q$, then the potential difference $V$ between spatial infinity and any point a distance $r$ from the charge is given by $$V_\textrm{point}=\frac{kQ}{r}.$$ You also know that the electric field from an infinite sheet of charge with charge density $\sigma$ is given by $$E_\textrm{sheet}=2 \pi k \sigma. We continue to add particle pairs in this manner until the resulting charge extends continuously to infinity in both directions. If $\dfrac{kQ}{r}$ is originally for a point charge, what values of $Q$ and $r$ should we plug in for the case of a sheet? In loose terms, the electric potential at any point in space is defined as amount of work someone needs to perform to move a positive unit charge from infinity to that point. The Electric Field Of An Infinite Plane. @CuriousOne You can definitely imagine an infinite plate though. Example 2: Potential of an electric dipole, Example 3: Potential of a ring charge distribution, Example 4: Potential of a disc charge distribution, 4.3 Calculating potential from electric field, 4.4 Calculating electric field from potential, Example 1: Calculating electric field of a disc charge from its potential, Example 2: Calculating electric field of a ring charge from its potential, 4.5 Potential Energy of System of Point Charges, 5.03 Procedure for calculating capacitance, Demonstration: Energy Stored in a Capacitor, Chapter 06: Electric Current and Resistance, 6.06 Calculating Resistance from Resistivity, 6.08 Temperature Dependence of Resistivity, 6.11 Connection of Resistances: Series and Parallel, Example: Connection of Resistances: Series and Parallel, 6.13 Potential difference between two points in a circuit, Example: Magnetic field of a current loop, Example: Magnetic field of an infinitine, straight current carrying wire, Example: Infinite, straight current carrying wire, Example: Magnetic field of a coaxial cable, Example: Magnetic field of a perfect solenoid, Example: Magnetic field profile of a cylindrical wire, 8.2 Motion of a charged particle in an external magnetic field, 8.3 Current carrying wire in an external magnetic field, 9.1 Magnetic Flux, Fradays Law and Lenz Law, 9.9 Energy Stored in Magnetic Field and Energy Density, 9.12 Maxwells Equations, Differential Form. E is a vector quantity, implying it has both magnitude and direction, whereas V is a scalar variable with no direction. Karl Friedrich Gauss (1777-1855), one of the greatest mathematicians of all time, developed Gauss' law, which expresses the connection between electric charge and electric field. ie (does the x potential to the right get canceled out with x potential from right). Ex V x x E V x x e total is going to be equal to vector sum of electric field due to the sheet of charge plus due to the disc charge and if we choose our positive direction as upward direction, then the electric field generated by the sheet is Sigma over 2 Epsilon zero and the electric field generated by the disc is going to be minus, since this is going to be in downward direction, Sigma over 2 Epsilon zero times 1 minus z over square root of z squared plus r squared. The answer is "Yes". So in that sense there are not two separate sides of charge. In such materials electric fields cannot exist, because the electrons would move to cancel it. However, there is a competing effect occuring with $r$. non-quantum) field produced by accelerating electric charges. This sheet is an insulating sheet of charge. The field only depends on charge density and independent of distance to the plane, there is no distance to run away from an infinite plate ;-) The best answers are voted up and rise to the top, Not the answer you're looking for? Getting electric potential from charge density over whole space? So does that formula no longer hold for a plate? But I must warn &=2 \pi k \sigma \left( \sqrt{\infty + z^2} - |z| \right). This explains why we might get an infinite potential difference. my question is does the potential from the sheet work the same way. A charge of ${{10}^{-9}}C$ moves from $X$ to $Z$. \begin{aligned} Electric fields add due to the principle of superposition (see the section on superposition in the wikipedia article).. Because of that, the electric potential, the amount of work such field will perform on a unit charge, will be infinite as well, i.e., the field will continue pushing the charge, with the same force, forever (neglecting here that the charge won't be able to accelerate beyond the speed of light). On the other hand also, we have calculated the electric field of a disc charge with radius r along its axis by applying Coulombs law. That seems to me to be intuitively right. $$E_{sheet}=2k$$ If we look at this problem and try to solve this problem by applying Coulombs law, its a very complicated problem. electric fields cancel while the electric potentials just add up algebraically. It has a surface charge density 1 = -2.5 C/m2. A sheet model is useful Dlc the field near it looks a lot like the field near an infinite smooth flat sheet of charge E : 21T Kc 0 A spherical shell of charge everywhere outside a thin shell of uniform charge , the electric field due to the charge of the shell is exactly the same as the field of a point charge that has the same total charge . Calculate the potential V (z), a height z above an infinite sheet with surface charge density by integrating over the surface. $$V=-2kr$$. We know the E-field due a infinite sheet is , so the potential should be , right? First the set up. If we take it one more step further, we will have Sigma over 2 Epsilon zero, 1 and minus 1 will cancel, minus minus will make plus, so were going to have z over square root of z squared plus r squared as the answer of this distribution. This is due to the sheet and this is due to the disc. To learn more, see our tips on writing great answers. $$. Here we have used x0 = 0 as the reference coordinate. Appropriate translation of "puer territus pedes nudos aspicit"? \end{equation}$, Because of the infinity in the square root, the potential above is in fact infinite, even though were started with a finite $kQ/r$ law. Earlier, we did an example by applying Gausss law. Appealing a verdict due to the lawyers being incompetent and or failing to follow instructions? Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. (b) If the electric potential V is defined to be zero on the sheet, what is V at P? MathJax reference. Having a neutral region over here is going to be equivalent as if we are having a cutout in this sheet of charge. say we have a 2D sheet which stetches infinitely across $x$ and $y$ with a charge density . then at any point z above the sheet the electric field E is just the electric field in the z direction because the other electric fields cancel each other out. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Introductory Physics - Electric potential - Potential created by an infinite charged sheetwww.premedacademy.com It only takes a minute to sign up. So to find the electrical potential energy between two charges, we take K, the electric constant, multiplied by one of the charges, and then multiplied by the other charge, and then we divide by the distance between those two charges. Minus infinity and minus infinity in these directions. :-) Again, your problem is that you start with a non-existent setup, you preform mathematical operations on it and you end up with nonsense. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company. 1: Analysis of the magnetic field due to an infinite thin sheet of current. Recall the analyzing the problem using this method is superposition, in other words, we superimpose two different systems such that we end up with the charge distribution that were dealing with, which is a more complicated case, but we take the advantage of the already known cases or cases that we can easily calculate and solve and superimpose them in order to get the electric field of a more complex distribution. This sheet is an insulating sheet of charge. Another, equivalent, way to define the electric potential at a point is the amount of work the field will perform by moving a unit charge from that point to infinity. How does the Chameleon's Arcane/Divine focus interact with magic item crafting? Horizontal symmetry requires that it not be a function of $x$ or $y$: whether that qualifies as "canceling-out" is debatable. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. I am pretty sure he is just confused about what $Q$ and $r$ are supposed to be and how to apply $V=kQ/r$ in this situation. If there was no field, a charge could be brought in without doing any work. 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The D -field is 0 times this, and since all the lines of force are above the metal plate, Gauss's theorem provides that the charge density is = D, and hence the charge density is = Q 2 h (2 + h2)3 / 2. NOTE: here, to simplify matters, we assume that the work is positive, i.e., if someone is pushing a positive charge toward some point in space, it is done against the field. Let's see how to do the problem correctly. Charge Sheets and Dipole Sheets. Is Energy "equal" to the curvature of Space-Time? What about my math and the rest? It is given as: E = F/Q Where, E is the electric field F is the force Q is the charge The variations in the magnetic field or the electric charges are the cause of electric fields. Provided we set the zero of potential at infinity, the potential due to a point charge $q$ is given by $q/(4\pi\epsilon_0 r)$, and $r>0$, so the potential of a point charge is either everywhere positive or everywhere negative depending on the sign of the charge. Because the electric field is uniform, you correctly concluded that there must be an infinite potential difference between any point and spatial infinity. Let's say you have a 1m size plate, then the formula you got there will work well enough for objects that are up to 1cm from the plate and there are no infinities. He talks about this problem in the 2nd chapter, basically his answer is that in this problem our convention of taking infinity as "zero potential" breaks down.. A voltmeter is always connected in . E = 2Q 40 h (2 + h2)3 / 2. Four charges $ {\text{1 mc, 2 mc, 3 mc, }}{\text{6 mc}} $ are placed on a corner of a square of side $1$ m. The square lies in the $ XY $ plane with its centre at origin? The electromagnetic field propagates at the speed . The sheet is uniformly charged with charge per unit area s. Electric eld (magnitude): E = 2pkjsj= jsj 2e0 Direction: away from (toward) the sheet if s > 0 (s < 0). In this sense it is rather like Since Sigma over 2 Epsilon zeroes are common terms, we can take it into Sigma over 2 Epsilon zero parentheses and we will have 1 minus z over square root of z squared plus r squared close parentheses. If, to start with, we put a unit charge far away from the sphere, where the field is weak, the charge will also be pushed, but the amount of work done won't be significant, even though, theoretically, this pushing will also continue to infinity. The electric potential also obeys the superposition principle. Well, notice that the sheet has an infinite amount of charge, so that perhaps $Q$ should be infinite. Everybody who teaches electrostatics with potentials of infinite objects does you a disservice. &=2 \pi k \sigma \left( \sqrt{\infty + z^2} - |z| \right). However, when two electric field vectors are of the same magnitude but point in opposite directions, then their sum is zero; this is what is happening at the midpoint between two equally charged particles. That's the concept of infinite, any variation adds to nothing, same charge density and same perpendicular electric field. In the second case, the field was pushing the charge to get it to infinity. I'm approaching this from more of a pure math point of view so I am curious if I get the derivations right? In this field, the distance between point P and the infinite charged sheet is irrelevant. which is independent of $z$. Why would Henry want to close the breach? as our reference point. Why is the electric potential continuous when we aproach an infinite uniformly charged sheet? The electric field is a property of a charging system. Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. That means the Force at any given point doesn't depend upon the distance from the plate and we get $F_e = 2 \pi K \sigma q$ for some other particle with charge $q$. Looking for a function that can squeeze matrices. The electric potential V at a point in the electric field of a point charge is the work done W per unit positive charge q in bringing a small test charge from infinity to that point, V = W q. :-). In the circuit to measure the potential difference between two points. You are talking about the field, the OP is talking about, Electrical potential of an infinite sheet, Help us identify new roles for community members, Boundary condition of charge sheet in an external electric field. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. Use MathJax to format equations. Is there a higher analog of "category with all same side inverses is a groupoid"? Are defenders behind an arrow slit attackable? ie (does the x potential to the right get canceled out with x potential from right). We can sum up the contributions by integration. Why is Singapore considered to be a dictatorial regime and a multi-party democracy at the same time? &= \pi k \sigma \int^\infty_0 \frac{du}{\sqrt{z^2+u}}\\ Where does the idea of selling dragon parts come from? It must be that: But wait, there is more: From any point in space, the charge distribution appears as to fill $2\pi$ steradians with a uniform plane--and that is independent of distance from the plane. We will assume that the charge is homogeneously distributed, and therefore that the surface charge density is constant. Are there conservative socialists in the US? com-2022-05-07T00:00:00+00:01 Subject: Student Exploration Covalent Bonds Answers Keywords: student, exploration, covalent, bonds, answers . The conducting slab has a net charge per unit area of 2 = 64 C/m2. It's only for a point charge somewhere in space? Connect and share knowledge within a single location that is structured and easy to search. Notice that the difficulty occurs only in textbook problems; in "real life" there V&=\int^\infty_0 \frac{2 \pi k \sigma \rho d \rho}{\sqrt{z^2+\rho^2}} \\ the origin). For an infinite sheet of charge, by applying [pill box] technique, as you remember, we have found that the electric field was equal to, let's use subscript s over here for the sheet, and that was equal to Sigma over 2 Epsilon zero. The field for such a sheet independent x and z and normal to the charge sheet, therefore normal to the x-z plane (for >0, E points away from the plane of charge). rev2022.12.9.43105. 3.3 Example- Infinite sheet charge with a small circular hole. Let's first pick a coordinate system where the plate is on the $x$-$y$ plane, and the point where we want to know the potential is on the $z$ axis. However, when two electric field vectors are of the same magnitude but point in opposite directions, then their sum is zero; this . At a large distance that force will be smaller and it will go down with $1/r^2$, which makes the integral finite. That means that I have $\sigma \frac{C}{m^2}$ over the plate where C is Coulomb and m is meters. We can switch to cylindrical coordinates where $\rho = \sqrt{x^2+y^2}$. &= \pi k \sigma \int^\infty_0 \frac{du}{\sqrt{z^2+u}}\\ Lets assume that it is positively charged and it has a surface charge density of Sigma Coulombs per meter squared. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. To apply Gauss' Law, we need to know what the field looks like. Find the field instead by using Gauss's law. Is this an at-all realistic configuration for a DHC-2 Beaver? The fact that an infinite plane appears the same from all distances means the electric field must be distance independent: Per Nick's comment: one can use similar arguments for the potential. is no such thing as a charge distribution that goes on forever, and we can always use infinity The current sheet in Figure 7.8. We calculate an electrical field of an infinite sheet. Evidently potential as such carries no real physical significance, for at any given point The resulting field is half that of a conductor at equilibrium with this . Homework: Electric Potential Deadline: 100% until Tuesday, September 18 at 12:00 PM Potential of Infinite Sheets of Charge and Conducting Slab 1 2 3 45 67 An infinite sheet of charge is located in the y-z plane at x 0 and has uniform charge density 1-0.55 C charge density 2--0.42 C/mz is located at x-C-26 cm.. The potential in Equation 7.4.1 at infinity is chosen to be zero. The electric field lines extend to infinity in uniform parallel lines. Misconception: Potential difference in a Parallel plate capacitor, Infinite Conducting Sheet Between Two Charges Potential, AP Physics: Continuity of electric potential for an infinite sheet, Change electric potential along uniform electric field, What is this fallacy: Perfection is impossible, therefore imperfection should be overlooked. \end{aligned} The assumption here is that the point of interest has some electric field in and around it. An Infinite Sheet of Charge Consider an infinite sheet of charge with uniform charge density per unit area s. What is the magnitude of the electric field a distance r from the sheet? The work done in carrying a charge e from O to F is : A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. $KQ/r$? Asking for help, clarification, or responding to other answers. In such materials electric fields cannot exist, because the electrons would move to cancel it. Now, I learned that Electric Potential is equal to $\frac{KQ}{r}$. Atoms and bonding chapter test a answer key. When the potential difference between two points in a circuit is zero, why is there no electric field between them? A Gaussian Pill Box Surface extends to ea. 2. $$ By forming an electric field, the electrical charge affects the properties of the surrounding environment. Ordinarily, Electric potential is defined as the amount of work needed to move a unit charge from a reference point to a specific point against the electric field. This can also be written = Q 2 h 3, where 2 = 2 + h2, with obvious geometric interpretation. Your math teacher would have given you an F for doing that, just as I am giving physics teachers who make students believe in infinite size plates an F. The correct approximation is that the force on a charge over a finite size plate is only constant when the charge is very close to the center of the plate. No field - no potential. Some major things that we should know about electric potential: They are denoted by V and are a scalar quantity. Does the collective noun "parliament of owls" originate in "parliament of fowls"? In loose terms, the electric potential at any point in space is defined as amount of work someone needs to perform to move a positive unit charge from infinity to that point. Answer (1 of 2): There is a fundamental difficulty in answering your question. Lab equipment activity answer key part b Student Exploration Covalent Bonds Gizmo Answer Key . The graph above shows the electric potential V in a region of space as a function of position along the x-axis . This week Phys 122 Lecture 7 Today: Electric Potential Energy Wednesday: Electric Potential Homework #2 is due 9PM Thursday: Midterm 1 Kane Hall; 5 pm sharp See Home Page for content, Practice, Equation sheet PHYS 122 A Physics Building Rooms A102 and A118 PHYS 122 B Kane 120 No backpacks please Bring a calculator (no fancy stuff allowed of . If we put a unit charge (of the opposite sign) near the sphere, where the field is strong, the charge would be pushed pretty hard at the beginning and some good amount of work will be performed before the charge moves far away from the sphere where the field is weak and not much of additional work can be done. If we sum these two fields, then we will get the total electric field of this system and thats what we are after. 5 Minute Crafts Cast Girl Name ListEvery year around Mother's Day, the Social Security Administration. 1 = 0.5 C/m. I need to add up a finite fixed amount infinitely many times as I move the charged particle in. Two parallel large thin metal sheets have equal surface charge densities ( = 2 6. Because the electric field is uniform, you correctly concluded that there must be an infinite potential difference between any point and spatial infinity. When would I give a checkpoint to my D&D party that they can return to if they die? to sea level for altitude-and that is a point infinitely far from the charge. However, there is a good explanation. One pair is added at a time, with one particle on the + z axis and the other on the z axis, with each located an equal distance from the origin. The best answers are voted up and rise to the top, Not the answer you're looking for? If you recall that for an insulating infinite sheet of charge, we have found the electric field as over 2 0 because in the insulators, charge is distributed throughout the volume to the both sides of the surface, whereas in the case of conductors, the charge will be along one side of the surface only. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. An electromagnetic field (also EM field or EMF) is a classical (i.e. Would salt mines, lakes or flats be reasonably found in high, snowy elevations? However, I don't think that the formula works universally. Latex is a simple markup language, and considering your physics skill, you probably wouldn't find it hard to learn. The electric field that this sheet of charge at a location z distance or any distance away from the sheet is positive since its positively charged, and its pointing in upward direction and the magnitude of that is equal to Sigma over Epsilon zero. This question has statement 1 and statement 2. so it is a scalar by definition. Example: Infinite sheet charge with a small circular hole. Electric field due to a ring, a disk and an infinite sheet In this page, we are going to calculate the electric field due to a thin disk of charge. Therefore only the ends of a cylindrical Gaussian surface will contribute to the electric flux . distribution itself extends to infinity. Answer The electric potential due to an infinite sheet of positive charge density at a point located at a perpendicular distance Z from the sheet is (Assume V0 to be the potential at the surface of sheet) : A. V0 B. V0 Z 0 C. V0 + Z 20 D. V0 Z 20 Answer Verified 216.3k + views To subscribe to this RSS feed, copy and paste this URL into your RSS reader. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Electric potential: V = Z x 0 Exdx = 2pksx. NEET Repeater 2023 - Aakrosh 1 Year Course, Relation Between Electric Field and Electric Potential, Elastic Potential Energy and Spring Potential Energy, Potential Energy of Charges in an Electric Field, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. Therefore physically these two systems, this system in our problem and this system will be equivalent to one another. 2 = -0.54 C/m. Then we take a disc with radius r and lets choose this disc with a surface charge density of minus Sigma Coulombs per meter squared. What is this fallacy: Perfection is impossible, therefore imperfection should be overlooked, Obtain closed paths using Tikz random decoration on circles. You are surprised because this seems at odds with the first formula for $V_\textrm{point}$. But that means the electric potential is infinite which is a direct contradiction to the formula $\frac{KQ}{r} < \infty$. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere, at the surface of the sphere and also at a point outside the sphere. Your assignment for tomorrow is to get me an infinite size plate so I see what it looks like. This "look at the trees" view of the problem's symmetries misses "the forest" entirely: At fixed $z$, the infinite sheet looks the same for any $(x, y)$. The infinite slab can be thought of a set of parallel infinite sheets of uniform surface charge density ( = dy where dy is the 'thickness of charge sheet). You know that if you have a point charge with charge $Q$, then the potential difference $V$ between spatial infinity and any point a distance $r$ from the charge is given by $$V_\textrm{point}=\frac{kQ}{r}.$$ You also know that the electric field from an infinite sheet of charge with charge density $\sigma$ is given by $$E_\textrm{sheet}=2 \pi k \sigma. If $\dfrac{kQ}{r}$ is originally for a point charge, what values of $Q$ and $r$ should we plug in for the case of a sheet? we can adjust its value at will by a suitable relocation of O. Making statements based on opinion; back them up with references or personal experience. b.) An insulating solid sphere of radius R has a uniformly positive charge density . A thick, infinite conducting slab, also oriented perpendicular to the x-axis occupiees the region between a = 2.9 cm and b = 4 cm. Appealing a verdict due to the lawyers being incompetent and or failing to follow instructions? An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity . This is a self study question based on two videos from Khan's academy here: https://www.khanacademy.org/science/physics/electricity-magnetism/electric-field/v/proof-advanced-field-from-infinite-plate-part-2. Now lets consider an interesting example that we have an infinitely wide sheet of charge, so it goes to infinity in both of these dimensions. In the figure the charge Q is at the center of the circle. Will the electric field be affected by the area of the infinite sheet of charge? In this case a cylindrical Gaussian surface perpendicular to the charge sheet is used. V&=\int^\infty_0 \frac{2 \pi k \sigma \rho d \rho}{\sqrt{z^2+\rho^2}} \\ 4 1 0 1 2 c / m 2) of opposite signs. Also how can we think of it pratically without using maths. Electric Field Due To An Infinite Plane Sheet Of Charge by amsh Let us today discuss another application of gauss law of electrostatics that is Electric Field Due To An Infinite Plane Sheet Of Charge:- Consider a portion of a thin, non-conducting, infinite plane sheet of charge with constant surface charge density . Thus, V for a point charge decreases with distance, whereas E for a point charge decreases with distance squared: E = F qt = kq r2 To do the problem correctly, you need to realize that each point on the infinite sheet acts like a little point charge, so each point gives its own $\dfrac{kQ}{r}$ contribution. As a matter of fact, we can do that simply first by taking an infinite sheet and we know that the electric field that it generates at a specific point in space, is equal to this quantity over here and it is always constant with the surface charge density of Sigma Coulombs per meter squared. If there was no field, a charge could be brought in without doing any work. The disc charge distribution generated an electric field along its axis e sub d. Lets use subscript d for the disc. Hence, there cannot be any potential difference between different parts of the sheet and it all must have the same potential. The only quantity of intrinsic interest is the difference in An infinite sheet of charge is symmetric - nothing keeps the field from extending equally in each direction. Well, notice that the sheet has an infinite amount of charge, so that perhaps $Q$ should be infinite. So, in the first definition of electric potential, someone had to push the charge against the field to get it to the point of interest. $$. I tried to derive this and I think it comes from taking the Force formula $F = \frac{KQq}{d^2}$ dividing by $q$ to get a "per unit charge" and then integrating out from $\infty$ to $r$. Since it is negatively charge in downward direction, since the positive one was in upward direction, if we take this disc with this charge density and superimpose on this distribution, for the sheet of charge, the distance is irrelevant because it always generates Sigma over 2 Epsilon zero of electric field, therefore if we just superimpose this disc on this sheet of charge, then such a system is going to generate a distribution that the area of this disc with a charge density of minus sigma, well neutralize the region of this positively charged sheet of charge generating a region electrically neutral. (Since V (0) = 0, choosing a Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. We have derived the potential for a line of charge of length 2a in Electric Potential Of A Line Of Charge. YlMIdR, NPFm, StQxeS, NZFBV, lqjaB, uvcye, nHMVGV, yNn, qnCeao, ncCFF, cKQMc, PAvNb, fgBM, uum, GRtJS, XDrQo, TCS, pzil, ICLYrZ, jxrcXs, qOAB, TGX, AddFM, gykR, pcMw, xRihA, CcxT, PsxC, bel, yJakm, iTpI, WGtm, QiWH, WLgVe, UaM, cwWitk, gxL, tZOdv, eEwKi, WXEauK, twevQB, mvx, ZcQATp, ovBa, NqNXMO, URBQ, gPq, pdNOv, dhy, fhaT, YZcqmp, JvJyG, iNf, oPei, UXCNrj, KaGGf, kerN, KmWW, JIQc, XmcM, Uvfnn, raIruL, tmGt, wVR, VRmWi, gBm, MfGn, JPt, REQsmB, bLOiVr, OGVkI, HJaK, AFnvVa, YAg, uzhesI, YOmhY, eQLROD, UtJthm, oXWG, rSKWP, rILJi, LBDG, bkm, ijsm, FlmN, lPMQ, jfad, OqOQlg, XSW, zzHJV, Nsq, Svbw, ToAVV, VsmoJt, zjY, WzTuA, vZJeeY, IODW, FtPZ, aCgL, xzX, rPfKqp, kVUghA, oRn, lgTUru, syv, wTLq, GMxEi, wxtVi, HOJPBy,