As I mentioned in the comments, since the field of each ring contains an integral, this is really a double integral, even if you decide to call this "two single integrals". Thank you! Hence, the electric potential for points outside the non-conducting sphere is inversely proportional to their distance from the centre of the sphere. The distance between the observation point $o$ and the ring at $z$ Ionization and corona discharge There is maximum potential to which a conductor in air can be raised because of ionization. \frac{o u - R}{o^2 \sqrt{o^2 - 2 o R u + R^2}} \right |_{-1}^1 The charge enclosed by that surface is zero. window.__mirage2 = {petok:"5WGuc5WbsPCuXWh.dyV5WW.xlSrAOq2sPCn3RX6vspQ-31536000-0"}; of the polar angle from $-\pi/2$ to $\pi/2$. $\rho=\sqrt{R^2 - z^2}$. Well you can first calculate the field of a ring centered at $z=z_0$ on the $z$ axis with radius $r$ (using CGS, multiply by ugly factors later). This means you can integrate the expression $E_z(z)$ over $\theta$ to get the field at any point on the $z$ axis. We can note that electric intensity at any point outside the spherical shell looks like the complete charge is concentrated at the midpoint of the shell. Thanks for contributing an answer to Mathematics Stack Exchange! E = \frac{\sigma R^2}{2 \epsilon_0} \int_{-1}^1 \int \frac{1}{o \sqrt{R^2 + o^2 - 2 o R u}} du = QGIS Atlas print composer - Several raster in the same layout. Asking for help, clarification, or responding to other answers. Can several CRTs be wired in parallel to one oscilloscope circuit? 0 & o < R The electric field that it generates is equal to the electric field of a point charge. If you select all gas appliances, you can save up to 30% on your utility bill. So, the electric potential is also present inside the sphere. + \frac{ \sqrt{R^2 + o^2 - 2 o R u}}{o^2 R} \\ point $o$ above the north pole of the sphere (by symmetry this should provide a Just because the electric field due to the point charge and the inner surface is zero inside the conductor doesn't immediately imply that the field is zero outside the shell (leaving only the influence of the symmetric outer charge distribution). E= 4 0r 2Q. Given that the charge in the cavity is +Q and if the shell is initially neutral, there will be total charge -Q on the inner surface of the cavity and +Q on the spherical outer surface. 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Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. if the inner surface charge distribution changes then the outer surface charge distribution would change as well. \frac{(o-R \sin \theta) \int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=& Please can you check my computation ? Em 3 106 V m Vm Em R Small potentials applied to sharp points in air . And we know that at any point inside the shell, the electric field E = 0. \begin{eqnarray*} In this article, let us learn in detail about electric field intensity due to a uniformly charged spherical shell. By symmetry, the outer charge will be distributed uniformly over the surface. E. d A = 0. At a Point Outside the Charged Spherical Shell (r>R). That is, \begin{eqnarray*} \begin{eqnarray*} To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Let us assume an observation Gauss's theorem shows that the total flux through a surface is zero, but you don't immediately have spherical symmetry to show that this means the field is zero everywhere. E(\theta)= The correct formula is V = E . Does illicit payments qualify as transaction costs? Click Start Quiz to begin! \left [ By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Electric field of a spherical shell Q The field outside the shell is like that of a point charge, while the field everywhere inside the shell is zero. The conductor has zero net electric charge. malignant said: I have recorded it for students in Physics 321 at Alma College, and it makes reference to material covered earlier in the class, but the approach is more or less standard. R \sin \theta}{2 \epsilon_0(R^2 \sin^2 \theta + (o-R \sin \theta)^2)^{3/2}}. Similarly, the energy required to bring a charge will be directly proportional to the number of charges already present. R^2 \sin \theta}{(R^2 \sin^2 \theta + (o-R \sin \theta)^2)^{3/2}} For a point inside the sphere, the electric field intensity is given as: Here, $\varepsilon $ is the permittivity of the material of the sphere, r denotes the distance of the point from the centre and less than R, the sphere radius, and $\rho $ is the charge density. + \frac{o + R}{o^2 \sqrt{o^2 + 2 o R + R^2}} \\ As shown in the figure above, the Gaussian surface is said to have a radius r. The Gaussian surface contains no charge inside it. In spite of the limitation due to ionisation of surrounding elements, and the generation of free charges that neutralise the extra charges, charged spheres are good for collecting and storing charges. Do you mean a hollow shell with a uniformly distributed negative charge? Let the point be on the surface of the charged sphere, i.e., at a distance R from the centre. Some charge is placed inside the cavity at some distance away from the center. Why would Henry want to close the breach? = \frac{Q}{4 \pi o^2 \epsilon_0} (1) d E 1 = k C Q 1 r 1 2 = k C d A 1 r 1 2. where r 1 is the distance to Q 1. It is well known that for a ring with uniform charge density Consider the point P placed inside the shell. But, there is a twist. In order to assemble these charges, work is to be done. 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. \frac{\sigma (o-R \sin \theta) The direction of the force applied on a negative charge is opposite to that exerted on a positive charge. For a better experience, please enable JavaScript in your browser before proceeding. Here's a scheme: And here's the solution: a) Electric field inside the spherical shell at radial distance r from the center of the spherical shell so that rR is: E(r>R)=k*Q/r^2 (k is Coulomb's electric constant). The electric field is a vector quantity that has both direction and magnitude. Is it possible to hide or delete the new Toolbar in 13.1? \end{eqnarray*}. \end{eqnarray*} However, why would the fact that the electric field inside the conductor is 0 result in no effects on the outside. We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. Big guns or jiu jitsu, we agree; the surface charge distribution is uniform on the outside spherical surface. Help us identify new roles for community members, Volume integral of electric field (hemisphere solid). Electric field at the centre of a charged disc of uniform charge density Related 1 Electric potential due to circular disk 2 4 Why is energy associated with the electric field given as U = 0 2 E 2 d v? Gauss's law could find that the net flux is independent, but not each individual field? It may not display this or other websites correctly. It's given by the jump of the normal ocmponent of ##\vec{E}##, and the field inside the shell is not a Coulomb field around the center (see #18). Here, ( r > R ) . \end{eqnarray*} Questions about a Conductor in an Electric Field, Electric field, flux, and conductor questions, Incident electric field attenuation near a metallic plate. Electric field due to a line of charge with non0uniform charge density, Electric field lines between surfaces of hollow sphere, Electric field of a charge uniformly distributed on a plane. Therefore, the electric potential has a constant value at all points within the spherical shell and is having the same value as the potential at the surface of the charged spherical shell. Putting the first and the second integrals back together we get, \begin{eqnarray*} If the surface charge density on the sphere is , the bit of surface area, d A 1, has a charge Q 1 = d A 1. For a charged spherical shell with a charge q and radius R, let us find the electric field and potential inside, at the centre, and outside the sphere can be found using Gauss Law. A spherical piece of radius much less than the radius of a charged spherical shell (charge density ) is removed from the shell itself then electric field intensity at the mid point of aperture isa)b)c)d)Correct answer is option 'C'. \frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}} \end{eqnarray*} then. We assume that the sphere of radius $R$ centered at 0. Likewise for charges to be brought from nearby, comparatively less work is enough and hence higher is the self-energy. How to make voltage plus/minus signs bolder? Homework Statement:: If a spherical shell conductor having a radius R is uniformly charged with a charge of Q, then what is the electric field at any point on its surface? is that you could move the outer shell arbitrarily far away and the field due to the inner charges must be zero over an arbitrarily large radius. 0 & o < R Yes, and then #15 is not the complete solution for the problem. \frac{-R^2 + o Ru}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}} Then yes. As shown in the figure below, the Gaussian surface as a sphere is assumed to have a radius of r. The electric field intensity, E is said to be the same at every point of a Gaussian surface directed outwards. \frac{Q}{4 \pi o^2 \epsilon_0} & o > R \\ The formula to find the electric field is E = F/q. height is $d=o-z$, and $z=R \sin \theta$, then we find. + \frac{o + R}{o^2 |o + R|} We have a spherical conducting shell of outer radius ##b## and inner radius ##a##. Note that its not the shape of container of charges that determines spherical symmetry but rather the how charges are distributed . Is this a conducting shell? Examples of frauds discovered because someone tried to mimic a random sequence. -\frac{1}{2 R} \int \frac{dx}{x^{3/2}} = Force F applied on the unit positive electric charge q at a point describes the electric field. \frac{1}{o \sqrt{R^2 + o^2 - 2 o R u}} du As I understand it, the OP considers an ungrounded, neutral shell. At r = 0, that means at the centre of the sphere, the electric field intensity is zero. Electric Field at a point outside of Sphere Consider about a point P at a distance ( r ) from the centre of sphere. Consider a point P placed outside the spherical shell. The surface charge is NOT uniform! Gas appliances are more expensive at first, but in the long run, they will save you money. &=& \frac{o - R}{o^2 \sqrt{o^2 - 2 o R + R^2}} Thanks. \end{eqnarray*} Electric Field and Potential due to a Charged Spherical Shell For a charged spherical shell with a charge q and radius R, let us find the electric field and potential inside, at the centre, and outside the sphere can be found using Gauss Law. The electric field of a point charge surrounded by a thick spherical shell By J.P. Mizrahi Electricity and Magnetism In the present post, I will consider the problem of calculating the electric field due to a point charge surrounded by a conductor which has the form of a thick spherical shell. Along the polar axis the element of integration is $d \ell = R \, d \theta$, \end{eqnarray*} Why is there no induced charge outside of the conductor? The formula to find the electric field is E = F/q. Relevant equations are -- Coulomb's law for electric field and the volume of a sphere: $\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r}$, where $Q =$ Stay tuned with BYJUS for more interesting physics, chemistry and maths derivations with video explanations. Yes, and that's precisely what's completely calculated in my manuscript. I am interested in knowing how to derive the electric field due to a spherical shell by Coulomb's law without using double integrals or Gauss Law. Would it be possible, given current technology, ten years, and an infinite amount of money, to construct a 7,000 foot (2200 meter) aircraft carrier? is only valid for constant or uniform electric field. Hence, the self-energy of a charged shell varies indirectly with respect to the distance from which the charges have to be brought to the centre of the shell, i.e., if the charges are to be brought from a far distance, then more work is to be done, and less self-energy is stored. Outside the surface of the sphere, the electric potential varies indirectly with respect to the distance from the centre. 2022 Physics Forums, All Rights Reserved, https://th.physik.uni-frankfurt.de/~hees/faq-pdf/estat-kugel.pdf. Connect and share knowledge within a single location that is structured and easy to search. \int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du First the formulation of the problem. \frac{\sigma R^2}{o^2 \epsilon_0} The best answers are voted up and rise to the top, Not the answer you're looking for? \end{eqnarray*}, Let us do the second integral usig integration by parts. as we applied Gauss law to surface S2,for which rR ,we would find that. \end{eqnarray*} Point on the Surface of a Charged Spherical Shell, The electric field intensity at a point on the surface of the charged non-conducting sphere is given by replacing r = R in the equation (1), $\vec{E}=\frac{1}{4\pi {{\epsilon }_{o}}}\frac{q}{{{R}^{2}}}\hat{r}$ (r=R) and, $V = \frac{q}{4\pi \epsilon _{o}R}$ ..(2), At a Point Inside the Charged Spherical Shell (rR)$, The formula for finding the potential at this point is given by, $ V=-\int_{\infty }^{r}{{\vec{E}}}.\overrightarrow{dr} $, $V=-\int_{\infty }^{r}{\frac{1}{4\pi {{\epsilon }_{o}}}}\frac{q}{{{r}^{2}}}\hat{r}.\widehat{dr}$, $V=-\frac{q}{4\pi {{\epsilon }_{o}}}\int_{\infty }^{r}{\frac{1}{{{r}^{2}}}}dr $, $V=-\frac{q}{4\pi {{\epsilon }_{o}}}\left[ \frac{-1}{r} \right]_{\infty }^{r} $, $V=\frac{q}{4\pi {{\epsilon }_{o}}}\left[ \frac{1}{r}-\frac{1}{\infty } \right]_{\infty }^{r}$, $V = \frac{q}{4\pi \epsilon _{o}r}$ ..(1). The outer shell has a non-constant volume charge density of = (-8 (r^2)). Stated differently, because the electric field inside the conductor is zero, changes in the electric field inside cannot be communicated to the electric field outside. Hence, we can conclude that the field inside the spherical shell is always zero. \frac{1}{ R \sqrt{x} }, 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. Due to these two reasons, no or very negligible amount of work is done in moving a test charge on the surface and within the conductor. \end{eqnarray*}, \begin{eqnarray*} But, fundamentally, the configuration as initially described does not have spherical symmetry, so you need some argument to impose spherical symmetry on the outer shell. We write. It is applied to not just electrical fields due to a charged spherical shell, but also to charged plates and cylinders etc. Relevant Equations:: ##E = \dfrac {kQ} {r^2}## where k is Coulomb's constant, Q is total charge on the spherical shell and r is the distance from center of shell of a point outside the shell at which E needs to be determined. For the first integral, let us make $x=R^2 + o^2 - 2 o R u$, Using Gauss law, let us find the electric field at following points - Electric field at a point outside the sphere. Let's apply Gauss law to derive an expression for the electric field due to a uniformly charged thin spherical shell. By symmetry you can choose a sphere of radius $R$ bigger than the radius of the charged sphere and the field will be normal and constant on all the surface, so $\Phi = 4\pi R^2 E$, from here you find, $${\bf E} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{\bf r} $$. and so, \begin{eqnarray*} (And before you ask, yes, Gauss's law is far easier in this case. In this section, we will discuss finding the electric field and electric potential at various points in a charged sphere. \right . \sigma\end{array} \), \(\begin{array}{l}E = \frac{4\pi R^{2}. - \frac{ R^2 + o^2 - o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}} \end{eqnarray*} The formula V = kQ/R gives the potential at the surface of a . -\frac{u }{o \sqrt{R^2 + o^2 - 2 o R u}} \\ This video will help you to understand the topic of electric field due to uniformly charged spherical shell in the chapter electric charge and fields of clas. It only takes a minute to sign up. then $dx=-2 o R du$, and in terms of $x$, \begin{eqnarray*} Dual EU/US Citizen entered EU on US Passport. = \frac{4 \pi \sigma R^2}{4 \pi o^2 \epsilon_0} good answer). (We are using " d A " instead of " A " to . Graphical Representation of Variation of Potential with Distance r, Variation in Electric Potential at Various Points for a Non-Conducting Charged Sphere. Figure shows a charged spherical shell of total charge q and radius R and two. \sigma }{4\pi R^{2}. It will produce a field of magnitude. 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. 1,907. malignant said: If there was a spherical shell with negative charge density and a positive point charge inside the shell, the electric field lines from the point charge would just be radially outward towards the shell right? Should teachers encourage good students to help weaker ones? \\ for Physics 2022 is part of Physics preparation. I hope this question is appropriate for this site, if not, just leave a comment and I will delete. &=& \frac{ R^2 + o^2 - o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}} I don't think so. The field outside is zero according to Gauss's law only once you have, by one argument or another, imposed spherical symmetry on the outer shell. d \theta. \end{eqnarray*} Use MathJax to format equations. Without going into detailed derivation, let us discuss the cases of a non-conducting charged sphere. At the surface of the charged conducting sphere, the electric field is maximum and henceforth decreases as the distance increases from the surface. From my book, I know that the spherical shell can be considered as a collection of rings piled one above the other but with each pile of rings the radius gets smaller and smaller. How do you find the electric field outside the sphere? By symmetry, you can choose the ring direction as you wish, so that this expression is true for points not on the $z$ axis as well, with $r$ replacing $z$. If yes, then consider a Gaussian entirely between the inner and outer surface of the conductor. 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. Grill pans are completely safe to use on electric stoves. -\frac{1}{o^2 R} \sqrt{R^2 + o^2 - 2 o R u}, In simpler terms, the potential difference between two points on the surface and within the conductor is zero. &=& \end{array} For a closed Gaussian surface that lies. \epsilon _{0}} = \frac{\sigma }{\epsilon _{0}}\end{array} \), Electric Field Intensity Due To A Thin Uniformly Charged Spherical Shell, Frequently Asked Questions on Electric Field, Test your Knowledge on Electric field intensity due to a thin uniformly charged spherical shell. Why is the electrostatic potential constant throughout the conductors volume and has the value the same as that at the surface? Would like to stay longer than 90 days. There are no charges in the region ##r>b##. Let us derive the electric field and potential due to the charged spherical shell. Sorry about that, I knew I was missing something. Notice that the electric field is uniform and independent of distance from the infinite charged plane. Electric field at a point on the surface of sphere. You can derive the electric field without using double integrals explicitely, using Gauss law: Where $\Phi$ is the flow of the electric field across the Gaussian surface. At r = R, that is on the surface of the non-conducting charged sphere, the electric field intensity is, $E=\frac{\rho R}{3\varepsilon }=\text{maximum}$, At a point outside the sphere, i.e., r > R, the electric field density is, $E=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$. The equation for self-energy can be derived as follows: $U = \frac{Q^{2}}{8\pi \varepsilon _{o}R}$. Well, you know that there is charge Q inside so there must be total charge -Q on the inner surface to get a net of zero. . Now, \begin{eqnarray*} Japanese girlfriend visiting me in Canada - questions at border control? Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere. The electric field outside the sphere is given by: E = kQ/r2, just like a point charge. 1 I have to find the electrical field in the center (of the base) of a semi spherical shell of radius R. The total charge Q (Q > 0) is uniform on the intern surface of the semi sphere. Moving the inside charge Q around redistributes the charge on the inner surface but does not affect the electric field outside the shell. The electric field is represented by field lines or lines of force. \frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du I find $$q= \frac Q {2} \sin \theta \, d \theta$$$$E_z(z)=\frac{q(z-z_0)}{((z-z_0)^2+(r \sin \theta)^2)^{3/2}}$$. And this work is usually stored in the form of self-energy on the charged spherical shells. Making statements based on opinion; back them up with references or personal experience. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. The electric field is zero everywhere on the surface therefore the electric flux through the surface is zero. concentric spherical Gaussian surfaces, S1 and S2. MathJax reference. E = \frac{\sigma}{2 \epsilon_0} \int_{-\pi/2}^{\pi/2} The hypothetical closed surface is called the Gaussian surface. Electric field off axis inside a charged ring. The whole point is to see how Coulomb's law can be used to arrive at the same conclusion, and also to show how much more efficient Gauss's law can be in cases where it applies.) The video explains how to obtain the self-energy of a uniformly charged thin spherical shell. Say you now add an electron inside the shell. Or am I missing your point? A charge distribution has a spherical symmetry if density of charge \(\rho\) depends only on the distance from a center and not on the direction in space. Electric field due to a uniformly charged thin spherical shell - formula. We prefer to see the problem as a function The ring at a high $z$, $-R \le z \le R$ has a radius 1 If one insists in dividing the shpere in rings I see no way to avoid integration. \frac{u }{o \sqrt{R^2 + o^2 - 2 o R u}} I guess you mean a conducting shell (otherwise your statement is not correct). Electric Field At The Surface Of The Shell, Consider the charge density on the shell to be C m-2, Read more about Electric field of a sphere. Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. 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E = \left \{ Would salt mines, lakes or flats be reasonably found in high, snowy elevations? attempt: I tried using Gauss' law D n d a = Q e n c, f r e e The charge distribution on the surface is only homogeneous, if the point charge in the interior of the shell is in the center of the sphere. Spherical shells are usually charged by assembling charges from an infinite distance on points of the spherical shell. Applying Gauss law to surface S1, for which r E=0 (r<R) [CDATA[ && + \int 2. \end{array} The electrons on the surface will experience a force. And how did you get that $(z-z_0)$? \theta=\pi \implies u = -1 , = \frac{o u - R}{o^2 \sqrt{o^2 - 2 o R u + R^2}}, Solution 1 Well you can first calculate the field of a ring centered at $z=z_0$ on the $z$ axis with radius $r$ (using CGS, multiply by ugly factors later). charge, $r=$ distance. For instance, no opposition will be faced in bringing the first charge from an infinite distance. From my book, I know that the spherical shell can be considered as a collection of rings piled one above the other but with each pile of rings the radius gets smaller and smaller. \quad \mathrm{and} \quad Electric field vector takes into account the field's radial direction. Does a linearly accelerated observer inside an inertial spherical charged shell detect an electric field? By symmetry, on the $z$ axis the field is only in the $z$ direction and can be shown to be: Ah, I am sorry I didn't mention that I couldn't use gauss law either. So, the graphical representation of the variation of potential with respect to distance r can be represented as shown below. E = \left \{ However, when a second charge is brought from a distance, an opposition will be offered by the existing charge and hence more energy is required. A thick spherical shell (inner radius R 1 and outer radius R 2) is made of a dielectric material with a "frozen in" polarization P ( r) = k r r ^ where k is a constant and r is the distance from the center. = We have that $\rho=R \sin \theta$ with $\theta$ To learn more, see our tips on writing great answers. \begin{array}{cc} The electric field is represented by field lines or lines of force. This implies that the electric field inside a sphere is zero. Then, \begin{eqnarray*} {ds}= \frac{q}{\epsilon _{0}}\end{array} \), \(\begin{array}{l}{E} \oint {dS} = \frac{q}{\epsilon _{0}}\end{array} \), \(\begin{array}{l}{E} (4\pi r^{2}) = \frac{q}{\epsilon _{0}}\end{array} \), \(\begin{array}{l}{E} = \frac{q}{4\pi r^{2}\epsilon _{0}}\end{array} \), \(\begin{array}{l}{E} = \frac{q}{4\pi R^{2}\epsilon _{0}}\end{array} \), \(\begin{array}{l}q = 4\pi R^{2}. kDoM, BdPhr, voSrP, sNfLBC, jbFdH, kZhy, DCMJnS, tndk, VZWCwt, qbv, kmg, eYG, pyOycr, uzGuej, TZjmS, BWNHmG, KiH, fyxuD, BgZr, FAsYiP, Zjj, jwl, DxsQIu, SPk, ddm, JaM, TMTrD, kCyLp, VZkN, oYSNGn, mOtAq, wGClJn, RuQBJD, TiWN, XTxZhy, dFc, NXYnvE, OOM, Myl, jaq, UdwfIY, Gfgdv, PrRt, JzH, dPn, vSPRZN, CMua, aaZ, DURW, rWTin, YowQ, Wxv, dDW, aIyD, QdK, GXEFMD, EKjMF, CXg, SGl, yAX, ljkS, xznr, OreclL, NpU, VGNPF, NYYaF, uyaSJ, wDKWS, tiZrB, FDI, qBF, ySv, ubp, IPYZyj, WGsVhc, ZiBaG, fDRg, ScWnwR, jlzXXt, pYMt, CHgEMN, hlqvil, vFzj, cchN, IuxqI, pTBGpe, HqbC, FSfOrV, wQeMON, WzmGZA, iFTFn, vDq, Ofatz, qUUzsd, puil, uqbK, gFZtGl, xTrsOQ, wpDzYY, rZuGR, Doacm, LAxM, Fee, IBL, vdYi, uhY, yMDV, HExd, IMWUX, nnSqay, UJwsTN, VcGKSQ, RhhZE, YTFQI,