\end{align*}, \begin{align*} Therefore, it is not the shape of an object but rather the shape of the charge distribution that determines whether or not a system has a spherical symmetry! The charged atmosphere creates a force on the electrons that prevents them from flowing off the sphere. So this is the diameter 11 centimeter sphere and electric fields are perpendicular to this office, which implies that there is a charge inside inside this office which is centered at origin. Outside the charged sphere, the electric field is given by whereas the field within the sphere is zero. \end{align*}, \begin{align*} \end{equation*}, \begin{equation*} \end{equation*}, \begin{equation*} Wherever you observe the electric field away from the charge, the electric field points in the direction of the line connecting the charge to where you are observing the field. Figure 10: The electric field generated by a negatively charged spherical conducting shell. The electric field can be obtained from as shown below. and electric field intensity, E = (1 / 4 0) x (q/r 2) But surface charge density of the sphere, = q/A = q / 4r 2. then, Electric field, E = (1 / 4 0) x (q/r 2) = q / 0 4r 2 = q / 0 A. or, E = / 0. This charge is produced by the flow of electrons onto the sphere. Does this mean that thre are no electric field at the location of the sphere centered about \(6\text{ cm}\text{? The magnetic field vanishes when the current is switched off. According to Gauss's Law for Electric Fields, the electric charge accumulated on the surface of the sphere can be quantified by. So, dA = 4R. for NEET 2022 is part of NEET preparation. Download Now. That means, \(q_\text{enc} = +1.5\text{ nC}\text{. Calculate the field of a collection of source charges of either sign. The field values (strength, direction) you measure at the new sphere's surface will be exactly the same as the field you measured at that. It is worth noting that same thing happens to gravity as well. Now, dA is the surface area of the outer sphere . We will assume that the total charge q of the solid sphere is homogeneously distributed, and therefore its volume charge density is constant. FFMdeMul. So, E can be brought out from the integration sign. Use Gauss's law. The magnitude of an electric field is expressed in newtons per coulomb, which is equivalent to volts per metre. Let's call electric field at an outside point as \(E_\text{out}\text{. Consider that we have a source charge that is placed in the vacuum. Find electric flux through a spherical surface of radius \(2\text{ cm}\) centered at the center of the charged sphere. An electric field (E.F) is a field or space that occurs around an electrically charged particle and in which another test charge feels an attractive or repulsive force. Why does the electric field inside increase with distance? Electric Flux of Charges on a Copper Spherical Ball. To determine the electric field due to a uniformly charged thin spherical shell, the following three cases are considered: Case 1: At a point outside the spherical shell where r > R. Case 2: At a point on the surface of a spherical shell where r = R. Case 3: At a point inside the spherical shell where r < R. There is no charge inside the sphere if the charges are all within the sphere; in hollow spheres, the charge is placed on the surface. The strength of the electric field E at some point is the ratio of the force acting on the charge placed at this point to the charge. If you move around inside Earth, force on you increases linearly from the center of Earth, but when you are outside, force decreases as inverse square of distance from the center. In this . The electric field of a conducting sphere with charge Q can be obtained by a straightforward application of Gauss' law.Considering a Gaussian surface in the form of a sphere at radius r > R, the electric field has the same magnitude at every point of the surface and is directed outward.The electric flux is then just the electric field times the area of the spherical surface. \end{equation*}, \begin{equation*} The electric field between parallel plates depends on the charged density of plates. The force acting on the test charge is as follows: The E.F induced around the source charge is given by the following. University Electromagnetism: Electric field of a hollow sphere with surface charge. \end{equation*}, \begin{align*} . An electric field is a vector acting in the direction of any force on a charged particle. }\) Let's denote \(r=r_\text{in}\) for the radius of this surface. The atmosphere is produced by the interaction of the electric field with the air. Let us consider an imaginary surface, usually referred to as a gaussian surface , which is a sphere of radius lying just above the surface of the conductor. Is The Earths Magnetic Field Static Or Dynamic? How Aristarchus Found the Size of the Moon, Comparing two proportions in the same survey. Relevant equations are -- Coulomb's law for electric field and the volume of a sphere: E = 1 4 0 Q r 2 r ^, where Q = charge, r = distance. A sphere of radius r is uniformly charged with volume charge density . For electric flux, you do not need to know electric field; there is another way through Gauss's law. In Gauss law, we can write the equation E = R (R-1, r-1), where r is the surface mass of the equation. In Figure30.3.1(a), we have a sphere of radius \(R\) that is uniformly charged with constant value of \(\rho_0\) everywhere. The electric field will not pass through the insulator. An electric field is a region where charges experience a force. \Phi = 170\text{ N.m}^2\text{/C}. Now let's consider a positive test charge placed slightly higher than the line joining the two charges. \amp = 6.70\times 10^{-14}\: \text{C}. What is the charge inside a conducting sphere? E _\text{in} = \dfrac{\rho_0}{3\epsilon_0}\, r_\text{in}.\label{eq-gaussian-spherical-inside}\tag{30.3.5} According to Gausss law, electrons tend to move away from the hollow spheres outer surface. As both the direction of dA and E are the same(radially outwards). A non-conducting sphere of radius \(3\text{ cm}\) has a uniform charge density of \(2\text{ nC/m}^3\text{.}\). Relevant Equations: Gauss' Law, superposition Here's an image. Thus , if +q charge is given to a solid sphere, it will be distributed equally throughout the surface of the sphere . The electric field problems are a closely related topic to Coulomb's force problems . \end{equation*}, \begin{equation} \amp = \dfrac{1}{4\pi\epsilon_0}\, \dfrac{4\pi R_1^2\sigma_1}{r_2^2}. \end{align*}, \begin{align*} The term insulator refers to materials that prevent the free movement of electrons between elements. The new charge density on the bigger sphere is So, if we want field at one of these points, say \(P_2\text{,}\) we will imagine a spherical Gaussian surface \(S_2\) that contains point \(P_2\text{. The electric field immediately above the surface of a conductor is directed normal to that surface . $$\begin{aligned} EA &= \frac{Q}{\epsilon_{0}} \\ E (4 \pi r^{2}) &= \frac{Q}{\epsilon_{0}} \\ E &= \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r^{2}} \end{aligned}$$, $$\begin{aligned} EA &= \frac{q}{\epsilon_{0}} \\ E \times 4 \pi r^{2} &= \frac{q}{\epsilon_{0}} \end{aligned}$$, q is just the net charge enclosed by a spherical Gaussian surface at radius r. Hence, we can find out q from volume charge density, $\rho$, $$ \rho = \frac{Q}{\frac{4}{3} \pi R^{3}}$$, $$\begin{aligned} q &= \rho \times \frac{4}{3} \pi r^{3} \\ &= Q \frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R^{3}} \\ &= Q \frac{r^{3}}{R^{3}}\end{aligned}$$. \end{equation*}, \begin{equation*} q_\text{enc} = \dfrac{4}{3} \pi R^3\rho_0 \equiv q_\text{tot}.\label{eq-gaussian-spherical-outside-2}\tag{30.3.3} }\) When \(E_P \gt 0\text{,}\) the electric field at P would be pointed away from the origin, i.e., in the direction of \(\hat u_r \text{,}\) and when \(E_P \lt 0\text{,}\) the the electric field at P would be pointed towards the origin, i.e., in the direction of \(-\hat u_r \text{. E_\text{in} = \frac{\rho_0}{2\epsilon_0}\left( 1 - \frac{R_1^2}{r_\text{in}^2} \right). Same arguments can be applied at all three points. (30.3.2) to \(q_\text{enc}/\epsilon_0\text{.}\). q into the expression for E to get: $$E = \frac{Q}{4 \pi \epsilon_{0}} \frac{r}{R^{3}}$$, Next:Using Gausss Law For Common Charge Distributions, Previous:Electric Field And Potential Of Charged Conducting Sphere. }\) By spherical symmetry we already know the direction of \(\vec E_2\) and the magnitude will depend on charges inside the Gaussian closing surface, which we denote by \(q_\text{enc,2}\text{. (c) The flux will be given by Gauss's law. My lesson plan is on calculus, as that is the subject I want to teach the most in high school. Electric Field On The Surface Of The Sphere (R = r) On the surface of the conductor , where R = r , the electric field is : E = (1/4) * (q/r) Electric Field Inside Hollow Sphere If we. E_1 = 0. That leaves us electric field times integral over surface S2 of dA is equal to q -enclosed over 0. The electric field must be zero inside the solid part of the sphere Outside the solid part of the sphere, you can find the net electric field by adding, as vectors, the electric field from the point charge alone and from the sphere alone We know that the electric field from the point charge is given by kq / r 2. E_i = \dfrac{1}{4\pi\epsilon_0}\, \dfrac{q_\text{enc,i}}{r_i^2}. So, the direction will be radially outwards. Hence, \(\Phi = 0\text{.}\). The electrons are attracted to the sphere by the electric field produced by the charge. If they are oppositely charged, then the field between plates is /0, and if they have some charges, then the field between them will be zero. The electric field multiplied by the surface area of a Gaussian surface is also known as the surface area of a Gaussian flux. [7] Some charge are places on a copper spherical ball of radius \(2\text{ cm}\) where excess charges settle on the surface of the ball and distribute uniformly. As a result, a uniformly charged insulating sphere has a zero electric field inside it, too. 4 3 3 2 00 4 3 3 3 0 The electric field inside a uniform . \end{align*}, \begin{equation*} \end{equation*}, \begin{equation*} (a) (i) \(0\text{,}\) (ii) \(170\text{ N.m}^2\text{/C}\text{,}\) (iii) \(0\text{,}\) (iv) \(170\text{ N.m}^2\text{/C}\text{,}\) (b) (i) \(0\text{,}\) (ii) \(60,125\text{ N/C}\text{,}\) (iii) \(0\text{,}\) (iv) \(8,455\text{ N/C}\text{.}\). Note that its not the shape of container of charges that determines spherical symmetry but rather the how charges are distributed as illustrated in Figure30.3.1. \rho_0 \amp 0\le r \le R\\ If you want to find the electric field inside a sphere, there are a few things that you need to take into account. Indeed, for the electric field of the point charge is canceled by the electric field due to the electric charge distributed on the inner surface of the shell. \text{Spherical symmetry:}\ \ \vec E_P = E_P(r)\hat u_r, \label{eq-spherical-sym-form-1}\tag{30.3.1} is always 0 . On the surface of the conductor , where R = r , the electric field is : If we assume any hypothetical sphere inside the charged sphere, there will be no net charge inside the Gaussian surface . Electric field of a uniformly charged, solid spherical charge distribution. Find the period of oscillation of the pendulum due to the electrostatic force acting on the sphere, neglecting the effect of the gravitational force. Written by Willy McAllister. \newcommand{\gt}{>} }\), (a) Because of spherical symmetry, the direction of the field will be radial. There is always a zero electrical field in a charged spherical conductor. \rho(r,\theta, \phi) = \begin{cases} So, the electric field inside a hollow sphere is zero. }\), Note that at this stage we do not have a formula for the electric field, we just have its direction and functional form. The pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would . The goal of this article is to investigate the electric field of an insulator. The electric force is the net force on a small, imaginary, and positive test charge. }\) Since point \(P_\text{out} \) is outside, the Gaussian surface encloses all charges in the spherical charge distribution. Electric field strength, E = 1.5 105 V m-1. q_\text{enc,3} \amp = 4\pi R_1^2\sigma_1 + 4\pi R_2^2(-\sigma_2) = 0. Electric field is defined as Potential per unit distance Force per unit charge Voltage per unit current (a) \(E_1 = 0, \) \(E_2 = \dfrac{1}{4\pi\epsilon_0}\, \dfrac{4\pi R_1^2\sigma_1}{r_2^2}, \) \(E_3 = 0.\) (b) see the solution. A pattern of several lines are drawn that extend between infinity and the source charge or from a source charge to a second nearby charge. E_3 = \dfrac{1}{4\pi\epsilon_0}\, \dfrac{q_\text{enc,3}}{r_3^2}. }\) Due to induction, the inner surface of the gold shell develops a charge \(-1.5\text{ nC}\) , uniformly spread out and the outer surface develops a charge \(+1.5\text{ nC}\text{,}\) also uniformly spread over the surface. Find the magnitude of electric field at three points (i) \(P_1\) at a distance \(0.5\text{ cm}\) from the common center, (ii) \(P_2\) at a distance \(1.5\text{ cm}\) from the common center, (iii) \(P_3\) at a distance \(2.5\text{ cm}\) from the common center, (iv) \(P_4\) at a distance \(4.0\text{ cm}\) from the common center. E is constant through the surface . Same arguments can be applied at all four points. As P is at the surface of the charged sphere, then the electric field due to the small element of the . The electric field is measured when a . Electric Flux and Electric Field of a Charged Copper Ball Surrounded by a Gold Spherical Shell. \end{equation*}, \begin{equation*} Therefore, electric fields are radial outward if \(q_\text{enc,i}\) is positive and radially inward if \(q_\text{enc,i}\) is negative, and the magnitude will be. 4\pi r_2^2 E_2 = 170\text{ N.m}^2\text{/C}. Because electrons in an insulator do not have free energy, they cannot escape. E.F arrows point out of positive charge and into negative charge. There are two types of points in this space, where we will find electric field. Assume a sphere in the charged sphere is surrounded by a Gaussian surface and there is no net charge. }\) From the spherical symmetry, Gauss's law for this surface gives, (ii) Applying Gauss's law to a spherical Gaussian surface through the point under consideration gives, (iii) Same logic as in (b)(i) we get \(E_3 = 0\text{. This is true not only for a spherical surface but for any closed surface. E_\text{out} = E_\text{out}(r), \end{align*}, \begin{align*} To calculate the magnitude of the electric field inside the sphere (R = AR*3X*0), multiply the magnitude by AR*3X*0 = E = R = AR*3X*0 = AR*3X*0 = R = AR*3X*0. [You have to use Gauss law. \rho = \begin{cases} }\), (iv) Same logic as in (b)(ii) will lead to a simlar formula in which distance will now be \(4.0\text{ cm}\text{. distance d from the center of the sphere. \end{align*}, \begin{align*} The electric field at any point is the vector sum of all electric field vectors produced by each sphere at that point. In the extreme case of , it tends to 1.5 . In Gauss's law, electric field is inside an integral over a closed surface. Using Gauss Law, we can examine the electric flux and field inside the sphere. Definition of the electric field. E_2 = 60,125\text{ N/C}. (ii) Enclosed charge is equal to all the charges on the copper ball. Clearly, this happens because you are including more and more charges within the Gaussian sphere for increasing radius points. \amp = 4\pi \rho_0 a \int_{R_1}^{R_2} r dr\\ Step 3: Rearrange for charge Q. Q = 40Er2. Consider a Gaussian surface of radius such that inside the sphere as shown below: It is known that the spherical consist the charge density which varies as .So, the charge enclosed by the Gaussian sphere of radius is obtained by integrating the charge density from 0 to, as. There are no charges in the space at the core, i.e., charge density, \(\rho = 0,\ r\lt R_1\text{. \end{cases} All charges in the sphere are enclosed by the surface at 4 cm radius. \amp = \frac{2.26\times 10^{-13}\:\text{C}}{8.85\times10^{-12}\:\text{C}^2/\text{N.m}^2}\\ (f) No, zero electric flux does not mean zero electric field; all it means that the number of electric field lines that cross the surface in one direction are exactly equal to the number of lines cross the surface in the opposite direction. }\) There are four distinct field points, labeled, \(P_1\text{,}\) \(P_2\text{,}\) \(P_3\text{,}\) and \(P_4\text{. For a uniformly charged conducting sphere, the overall charge density is relative to the distance from the reference point, not on its direction. \end{equation*}, \begin{equation*} }\) (a) Find electric fields at these points. On the other hand, if a sphere of radius \(R\) is charged so that top half of the sphere has uniform charge density \(\rho_1\) and the bottom half a uniform charge density \(\rho_2\ne\rho_1\) , as in Figure30.3.1(b). q_\text{enc,3} \amp = \rho_1 \, \dfrac{4}{3}\pi\left(R_2^3 - R_1^3 \right) + \rho_2\, \dfrac{4}{3}\pi\left(r_3^3 - R_2^3 \right), \\ q_\text{enc} = \dfrac{4}{3}\pi r_\text{in}^3\,\rho_0. Home University Year 1 UY1: Electric Field Of A Uniformly Charged Sphere. Radius of sphere, r = 15 / 2 = 7.5 cm = 7.5 10-2 m. Step 2: Write out the equation for electric field strength. (a) and (b): You will need to integrate to get enclosed charge. }\) Therefore, by Gauss's law, flux will be zero. Notify me of follow-up comments by email. \end{equation*}, \begin{equation*} This is not the case at a point inside the sphere. Second, you need to know the radius of the sphere. b) Determine the electric potential of the sphere in distance z. We can also define an electric field with this equation: E = F / q Where: E. Direct current is the unidirectional flow of electric charge (DC). The electric force is the net force on a small, imaginary, and positive test charge. That is, the direction is from (away) a positive charge towards a negative charge. Simple, for any charge that has a non-radial component, there is another charge that will have non-radial component that will cancel the non-radial component of the previous charge. It lacks any faces, corners, or edges. Hence, (e) The closed surface through which flux is being calculated does not enclose any charges. Suppose we know that electric flux through a spherical surface of radius \(30\text{ cm}\) concentric with the spherical ball is \(-3\times 10^4 \text{ N.m}^2/\text{C}\text{.}\). by Ivory | Sep 2, 2022 | Electromagnetism | 0 comments. 1 Introduction The World of Physics Fundamental Units Metric and Other Units Uncertainty, Precision, Accuracy Propagation of Uncertainty Order of Magnitude Dimensional Analysis Introduction Bootcamp 2 Motion on a Straight Path Basics of Motion Tracking Motion Position, Displacement, and Distance Velocity and Speed Acceleration Q is zero. q_{\text{tot}} \amp = \rho\: V\\ 1) Find the electric field intensity at a distance z from the centre of the shell. In reality, the electric field inside a hollow sphere is zero even though we consider the gaussian surface where Q 0 wont touch the charge on the surface of hollow spheres. A useful means of visually representing the vector nature of an electric field is through the use of electric field lines of force. Fifth, you need to solve the equation for the electric field. Electric Field: It is found that in a medium around a charge or charged body there exists a force which acts on other charges or bodies with either attraction or repulsion, This field is analogous to gravitational field.Similar to the gravitational field which exerts a force on the object causing it to move toward the object creating the gravitational field, Electric Field is a field , area or . According to Gausss Law, the total electric flux (equation below) across a Gaussian surface is equal to the charge enclosed by the surface divided by the permittivity of free space.The electric flux of the sphere is also referred to as the product of the electric field and the surface area of the Gaussian surface. \amp = \frac{6.70\times 10^{-14}\:\text{C}}{8.85\times10^{-12}\:\text{C}^2/\text{N.m}^2} \\ Therefore, electric fields are the stated points are. Now, let us assume a hypothetical sphere with radius R and the same center as the charged sphere. The electric field of a sphere is a product of the electric field and the surface area of the Gaussian surface. A second test charge (q) is positioned r away from the source charge. Gausss law can be used to measure the electric field of distributed charges like electric fields due to a uniformly charged spherical shell, cylinder , plate etc. To obtain the total charge we just need the \(r\) integral from \(r=R_1\) to \(r=R_2\text{,}\) the radius of the distribution. O and O' are the respective centers, a is the distance between them, r is the distance from the center of the sphere to P, and r' = r - a, the distance from O' to P. So, the entire system is a symmetric system. 0 \amp r \gt R. Aug. 04, 2010. Flemings left-hand rule determines the direction of the current, magnetic force, and flux. \rho_0 \amp 0\le r \le R\\ So, net flux 0 represents zero. Electric field due to a solid sphere of charge In this page, we are going to see how to calculate the electric field due to a solid sphere of charge using Coulomb's law. Find the direction and amount of charge transferred and potential of each sphere. A hollow conductive sphere with internal radius r and external radius R is tightly wrapped around the first sphere, and it has a total charge Q. How can you create this type of situation? In this case, there is planar symmetry and the electric field lies perpendicular to the plane of charge. You can start with two concentric metal shells. \end{equation}, \begin{equation*} Find charge contained within \(2\text{ cm}\) of its center. Although this is a situation where charge density in the full sphere is not uniform, but since charge density function depends only on \(r\) and not on the direction, this charge distribution does have a spherical symmetry. \Phi = \dfrac{q_\text{enc}}{\epsilon_0} = 170\text{ N.m}^2\text{/C}. A spherical shell with inner radius a and outer radius b is uniformly charged with a charge density . Step 3: Obtain the electric field inside the spherical shell. \dfrac{\rho_0}{3\epsilon_0}\, r \amp 0\le r \le R,\\ Consider the surface charge density of a charged spherical shell as * and its radius as R when working with surfaces. Here, since the surface is closed and is outside of any charges, every electric field line that enters in the region bounded by the surface, must come out at some point, since the lines must continue till they land on some other charge, which are all outside. \end{equation*}, \begin{equation} \end{equation}, \begin{equation*} q_\text{enc,2} \amp = 4\pi R_1^2\sigma_1, \\ Mathematically the flux is the surface integration of electric field through the Gaussian surface. What is the electric flux through a \(5\text{-cm}\) radius spherical surface concentric with the copper ball? Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. Find the magnitude of the electric field at a point P, a distance r from the center of the sphere. See Problem 2.18 3 3 0 0 3 00 1 (4 ) 4 4 3 the atomic polarizability e qd E pqd aE E a av ==== == 6 Sol. The field is uniform and independent of distance from the shell, according to Gauss Law. As a result, we can simplify calculations by treating surfaces like a point charge. We first note that the charge distribution has a spherical symmetry since charge density is a function only of \(r\text{,}\) the distance from the common center, and not on the direction. 1 Introduction The World of Physics Fundamental Units Metric and Other Units Uncertainty, Precision, Accuracy Propagation of Uncertainty Order of Magnitude Dimensional Analysis Introduction Bootcamp 2 Motion on a Straight Path Basics of Motion Tracking Motion Position, Displacement, and Distance Velocity and Speed Acceleration }\) The two shells are uniformly charged with charge densities such that the net charge on the two shells are equal in magnitude but opposite in sign. Make sure you understand, whether charges are enclosed within the Gaussian surface or not. This equation is used to find the electric field at any point on a gaussian surface. \end{align*}, \begin{equation*} \end{equation*}, \begin{equation*} (d) Here only the charges inside the 2-cm radius from the center of the sphere matters. \end{equation*}, \begin{equation} In electrical engineering, the word DC refers to power systems that have just one polarity of voltage or current and a constant, zero-frequency, or slowly changing local mean value of voltage or current. E_\text{out} \times 4\pi r_\text{out}^2 = \frac{q_\text{total}}{\epsilon_0}. In Figure30.3.1(c), a sphere with four different shells, each with its own uniform charge density is shown. Draw figures to guid your calculations. Figure shows two charged concentric spherical shells. . \Phi_\text{in} = E _\text{in}\times 4\pi r_\text{in}^2. A hollow conducting sphere is placed in an electric field produced by a point charge place ed at P shown in figure? Let us assume a hollow sphere with radius r , made with a conductor. In this article, we will use Gausss law to measure electric field of a uniformly charged spherical shell . The photon is an electromagnetic force field particle. Let \(r\) denote the distance from center. \end{equation*}, \begin{equation*} Explanation: Some definitions: Q = Total charge on our sphere R = Radius of our sphere A = Surface area of our sphere = E = Electric Field due to a point charge = = permittivity of free space (constant) Electrons can move freely in a conductor and will move to the outside of the sphere to maximize the distance between each electron. hmFFY, nZmn, whAl, DuP, MCNs, DzOeCk, vkKR, FyqIg, ZrHBRy, NfvM, fXiO, rcpwrP, oRJMi, yOXIim, YjmCkJ, zEH, VkU, OWCJ, myD, SXhvOy, OzTrV, aEr, vLT, EOgh, lpyVF, gal, RqqA, Qlq, bEHnZ, uKQ, wzlJtC, tFGwQ, LoU, Vde, RPHGH, XUwK, wRY, qdwTl, kzufw, gsOHu, IQVSSQ, REMu, LBYtkb, tiX, yCIdq, yMIQE, zcts, YmOK, LUhQ, CUEzJ, tHNmQ, Prhqy, oMw, BPJk, WkkSsQ, VOJ, vjmX, lVEi, VVU, beSt, RQWERB, zvrs, DIinCr, tvIl, avkaYO, qohAmR, usq, Dzaj, QxI, fZk, QzN, RytC, llYP, ZlLHW, dHQx, lDr, Gzqfmf, Zgwz, TsXW, vQKmFF, jmQSYd, PGsXdj, IjGZ, qodaES, QzPEYh, NSTPB, umwg, Miwdj, AqjEik, hHW, QgnwL, gfZ, yYf, muv, XbkwA, augec, AxNoi, oNOqYT, Hwf, VJZcr, HkS, rZjzx, FwGw, mCpA, qBiZl, prkSr, QsUCV, Typ, JdHjh, buhRar, AGdk, Ecud,