how was gauss's law derived

The Lagrangian density for Newtonian gravity is. Experimental Proof. The standard examples for which Gauss' law is often applied are spherical conductors, parallel-plate capacitors, and coaxial cylinders, although there are many other neat and interesting charges configurations as well. It is represented as. 0 Gausss law for electricity states that the electric flux across any closed surface is proportional to the net electric charge q enclosed by the surface; that is, = q/0, where 0 is the electric permittivity of free space and has a value of 8.854 1012 square coulombs per newton per square metre. V The entry in Wikipedia for Divergence Theorem says it was discovered by. This is because both Newton's law and Coulomb's law describe inverse-square interaction in a 3-dimensional space. To find the value of q, consider the field at point P inside plate A. G R Since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. be the electric field created inside and outside the sphere respectively. The electric field is the basic concept of knowing about electricity. the resultant field is that of all masses not including the sphere, which can be inside and outside the sphere). A closed surface is one that divides the universe up into two parts: inside the surface, and, outside the surface. List the three branches of the government and types of law that each creates. These electric field lines will extend to infinity decreasing in strength by a factor of one over the distance from the source of the charge squared. In physics, Gauss's law for gravity, also known as Gauss's flux theorem for gravity, is a law of physics that is equivalent to Newton's law of universal gravitation. The Gauss law is interpreted in terms of the electric flux through the surface. Abstract. For analogous laws concerning different fields, see, Deriving Newton's law from Gauss's law and irrotationality, Poisson's equation and gravitational potential, Cylindrically symmetric mass distribution, Del in cylindrical and spherical coordinates, The mechanics problem solver, by Fogiel, pp535536, Degenerate Higher-Order Scalar-Tensor theories, https://en.wikipedia.org/w/index.php?title=Gauss%27s_law_for_gravity&oldid=1155115481, All Wikipedia articles written in American English, Creative Commons Attribution-ShareAlike License 3.0. In equation form, Gausss Law reads: \[\oint \vec{E} \cdot \vec{dA}=\frac{Q_{\mbox{ENCLOSED}}}{\epsilon_0}\label{33-1} \]. I choose one that passes through both the point charge, and, point \(P\). Omissions? All in all, we can determine the relation between the Gauss law and Coulombs law by deducing the spherical symmetry of the electric field and by performing the integration. From Coulomb's Law, the . It may look more familiar to you if we write it in terms of the Coulomb constant \(k=\frac{1}{4\pi\epsilon_o}\) in which case our result for the outward electric field appears as: Its clear that, by means of our first example of Gausss Law, we have derived something that you already know, the electric field due to a point charge. In the case of a charged conducting sphere, So, if a and b are the radii of a sphere and spherical shell, respectively, the potential at their surfaces will be. It is negative when \(q\) is negative. B Thus, Q1 q = (Q1 + Q2)/2 . For analogous laws concerning different fields, see, Equivalence of integral and differential forms, Equivalence of total and free charge statements, More specifically, the infinitesimal area is thought of as, Uniqueness theorem for Poisson's equation, "Sur l'attraction des sphrodes elliptiques", "On the Covariant Representation of Integral Equations of the Electromagnetic Field", MIT Video Lecture Series (30 x 50 minute lectures)- Electricity and Magnetism, section on Gauss's law in an online textbook, https://en.wikipedia.org/w/index.php?title=Gauss%27s_law&oldid=1148160609, This page was last edited on 4 April 2023, at 12:50. = V {\displaystyle \mathbf {g} \cdot d\mathbf {A} =-4\pi GM}. Gausss Law in the form \(\Phi_E=\frac{Q_{\mbox{ENCLOSED}}}{\epsilon_0}\) makes it easy to calculate the net outward flux through a closed surface that encloses a known amount of charge \(Q_{\mbox{ENCLOSED}}\). The Gauss theorem statement also gives an important corollary: The electric flux from any closed surface is only due to the sources (positive charges) and sinks (negative charges) of the electric fields enclosed by the surface. The following diagram might make our conceptual statement of Gausss Law seem like plain old common sense to you: The closed surface has the shape of an egg shell. Relation to Gauss's law Deriving Gauss's law from Coulomb's law. Under these circumstances, Gauss's law modifies to. A Where is the linear charge density. This results in positive charges causing a positive flux and negative charges creating a negative flux. e principles of physical science: Gausss theorem, This article was most recently revised and updated by, https://www.britannica.com/science/Gausss-law, University of Saskatchewan - Introduction to Electricity, Magnetism, and Circuits - Explaining Gausss Law. Therefore the flux through a closed surface generated by some charge density outside (the surface) is null. One of the fundamental relationships between the two laws is that the Gauss law can be used to derive Coulombs law and vice versa. r ( Unless explicitly set forth in the applicable Credits section of a lecture, third-party content is not covered under the Creative Commons license. Well, there is a Gauss's Law for the magnetic field as well. 2. Pillbox, when the charge distribution has translational symmetry along a plane. Strictly speaking, Gauss's law cannot be derived from Coulomb's law alone, since Coulomb's law gives the electric field due to an individual, electrostatic point charge only. A closed surface is one that divides the universe up into two parts: inside the surface, and, outside the surface. By comparing equation (1) and (2) ,we get. Using the equation E = /20, the electric field at P. The net electric field at P due to all the four charged surfaces is (in the downward direction). Question: Use Gauss's Law to derive an expression for the electric field at a distance r from an infinite line charge with charge density A. Consider a point charge. {\displaystyle \mathbf {r} _{0}\in \Omega } The circle on the integral sign, combined with the fact that the infinitesimal in the integrand is an area element, means that the integral is over a closed surface. Explain all details necessary for using Gauss's Law to solve this problem. {\displaystyle e(\mathbf {r,\mathbf {r} '} )\in C^{1}(V\times \Omega )} Indeed, the constant of proportionality has been established to be \(\frac{1}{\epsilon_0}\) where \(\epsilon_0\) (epsilon zero) is the universal constant known as the electric permittivity of free space. At this point we need to choose a Gaussian surface. , In the case of an infinite line of charge, at a distance, r. The inner surface of C must have a charge -q from Gauss law. E ), and Poisson's equation becomes (see Del in cylindrical and spherical coordinates): When solving the equation it should be taken into account that in the case of finite densities /r has to be continuous at boundaries (discontinuities of the density), and zero for r = 0. In the case of the divergence theorem, only dates are given, no references. 0 R r In one sense, it is quite similar because it involves a quantity called the magnetic flux which is expressed mathematically as B = B d A and represents the number of magnetic field lines poking outward through a closed surface. 93 1 4 I think your broad question is: how to know that a contained charge will create a particular E. field. {\displaystyle \rho } Gauss's law, either of two statements describing electric and magnetic fluxes.Gauss's law for electricity states that the electric flux across any closed surface is proportional to the net electric charge q enclosed by the surface; that is, = q/ 0, where 0 is the electric permittivity of free space and has a value of 8.854 10 -12 square coulombs per newton per square metre. Imagine one in the shape of a tin can, a closed jar with its lid on, or a closed box. It will balance the weight of the particle, if, q 2.26 105N/C = 5 10-9 kg 9.8 m/s2, or, q = [4.9 10-8]/[2.26 105]C = 2.21 10-13C, The charge on one electron is 1.6 10-19C. Furthermore, the magnitude of the electric field has to have the same value at every point on the shell. The larger the number of field lines emanating from a charge the larger the magnitude of the charge is, and the closer together the field lines are the greater the magnitude of the electric field. At the time, we stated that the Coulomb constant \(k\) is often expressed as \(\frac{1}{4 \pi \epsilon_0}\). In fact, if I assume the electric field at any point \(P\) in space other than the point at which the charge is, to have a tangential component, then, I can adopt a viewpoint from which point \(P\) appears to be to the right of the charge, and, the electric field appears to be upward. such that We first prove that the electric field due to a point charge can have no tangential component by assuming that it does have a tangential component and showing that this leads to a contradiction. Before we consider that one, however, lets take up the case of the simplest charge distribution of them all, a point charge. (Q1 q)/2A0 q/2A0+ q/2A0 (Q2 + q)/2A0. At the given area, the field is along the Z-axis. = G Problem 7: A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. When the charge is uniformly distributed over the surface of the conductor, it is called surface charge density. d Thus, the angle between the area vector and the electric field is 90 degrees, and cos = 0. Now lets decide on a rotation axis for testing whether the electric field is symmetric with respect to rotation. The gravitational field g (also called gravitational acceleration) is a vector field a vector at each point of space (and time). ( Using Gauss law, the total charge enclosed must be zero. It is mathematically identical to the proof of Gauss's law (in electrostatics) starting from Coulomb's law.[2]. Therefore, we have derived Gauss's law: the total electric flux through a closed surface is equal to the total charge inside the surface divided by 0. ) Now consider So, for the case at hand, Gausss Law takes on the form: \[\oint E dA=\frac{Q_{\mbox{enclosed}}}{\epsilon_o} \nonumber \]. The next step involves choosing a correct Gaussian surface with the same symmetry as the charge distribution. Its consequences should also be identified. As the electric field in a conducting material is zero, the flux. This yields: \[E 4\pi r^2=\frac{q}{\epsilon_o} \nonumber \], \[E=\frac{1}{4\pi\epsilon_o} \frac{q}{r^2} \nonumber \]. We can choose the size of the surface depending on where we want to calculate the field. Gauss law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. If it were different at a point \(P\) on the spherical shell than it is at a point \(P\) on the spherical shell, then we could rotate the charge distribution about an axis through the point charge in such a manner as to bring the original electric field at point \(P\) to position \(P\). The distribution should be like the one shown in figure (b). For a given \(E\) and a given amount of area, this yields a maximum value for the case of \(\theta=0^{\circ}\) (when \(\vec{E}\) is parallel to \(\vec{dA}\) meaning that \(\vec{E}\) is perpendicular to the surface); zero when \(\theta=90^{\circ}\) (when \(\vec{E}\) is perpendicular to \(\vec{dA}\) meaning that \(\vec{E}\) is parallel to the surface); and; a negative value when \(\theta\) is greater than \(90^{\circ}\) (with \(180^{\circ}\) being the greatest value of \(\theta\) possible, the angle at which \(\vec{E}\) is again perpendicular to the surface, but, in this case, into the surface.). V In the case of an infinite uniform (in z) cylindrically symmetric mass distribution we can conclude (by using a cylindrical Gaussian surface) that the field strength at a distance r from the center is inward with a magnitude of 2G/r times the total mass per unit length at a smaller distance (from the axis), regardless of any masses at a larger distance. We can further say that Coulombs law is equivalent to the Gauss law, meaning they are almost the same. For example, a point charge q is placed inside a cube of the edge a. 4 The field between two parallel plates of a condenser is E =/0, where is the surface charge density. Our editors will review what youve submitted and determine whether to revise the article. (Youve seen \(\epsilon_0\) before. How many electrons are to be removed to give this charge? If no charges are enclosed by a surface, then the net electric flux remains zero. 3. More generally, for a mass distribution with the density depending on one Cartesian coordinate z only, gravity for any z is 2G times the difference in mass per unit area on either side of this z value. We can obtain an expression for the electric field surrounding the charge. But the total charge given to this hollow sphere is 6 10-8 C. Hence, the charge on the outer surface will be 10 10-8C. It is given by E = q enclosed 0, where E is the electric flux through the gaussian surface generated by the group of charges measured in V m, q enclosed is the total enclosed charge by the gaussian surface measured in coulombs C, and 0 is the vacuum permittivity given by a value of 8.85 10 12 F m. Due to the charge -q on the inner surface of B= -q/4, Due to the charge q on the outer surface of B =q/4, Due to the charge -q, on the inner surface of C =-q/4, Due to the charge q q on the outer surface of C = (q q)/4. r {\displaystyle \mathbf {r} _{0}} If we take the Gauss law, it is represented as, Meanwhile, the electric flux E can now be defined as a surface integral of the electric field. Define the term "separation of powers." 4. r Mathematical formulations for these two lawstogether with Ampres law (concerning the magnetic effect of a changing electric field or current) and Faradays law of induction (concerning the electric effect of a changing magnetic field)are collected in a set that is known as Maxwells equations, which provide the foundation of unified electromagnetic theory. The area = r2 = 3.14 1 cm2= 3.14 10-4 m2. To get a feel for what to expect, let's calculate the electric flux through a spherical surface around a positive point charge q, since we already know the electric field in such a situation. 1 Answer Aritra G. Mar 25, 2017 I shall outline a better method. In such cases, the right choice of the Gaussian surface makes \(E\) a constant at all points on each of several surface pieces, and in some cases, zero on other surface pieces. having Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. What will be the new potential difference between the same two surfaces, if the shell is given a charge -3Q? To further exploit the symmetry of the charge distribution, we choose a Gaussian surface with spherical symmetry. B This is because by the presence of charge on the outer shell, potential everywhere inside and on the surface of the shell will change by the same amount, and hence, the potential difference between sphere and shell will remain unchanged. r Almost any will do. {\displaystyle R} In the course for which this book is written, you will be using it in a limited manner consistent with the mathematical prerequisites and co-requisites for the course. {\displaystyle \mathbf {E} _{C}} | This means that the number of electric field lines entering the surface equals the field lines leaving the surface. In fact, Gauss's law does hold for moving charges, and in this respect Gauss's law is more general than Coulomb's law. The form of Gauss's law for gravity is mathematically similar to Gauss's law for electrostatics, one of Maxwell's equations. ) Introduction Physical Properties Maxwell I: Fundamentals Formative Laws Gauss's Law for Electric Fields Gauss's Law from Coulomb's Law Gauss's Law for Magnetic Fields Integral equation Differential equation Derivation using Biot-Savart law Differential equation in the frequency-domain Units Discoverers of the law Faraday's Law Ampere-Maxwell In the case of a spherically symmetric mass distribution we can conclude (by using a spherical Gaussian surface) that the field strength at a distance r from the center is inward with a magnitude of G/r2 times only the total mass within a smaller distance than r. All the mass at a greater distance than r from the center has no resultant effect. Find the flux of the electric field through a circular area of radius 1 cm lying completely in the region where x, y and z are all positive and with its normal, making an angle of 600 with the Z-axis. Thus, at each point in space, the electric field must be either directly toward the point charge or directly away from it. Now, if we apply Coulombs law, the electric field generated is given by, Where k=1 /40. But this would represent a change in the electric field at point \(P\), due to the rotation, in violation of the fact that a point charge has spherical symmetry. {\displaystyle \mathbf {r} \neq \mathbf {r'} } Also, if a given electric field line pokes through the surface at more than one location, you have to count each and every penetration of the surface as another field line poking through the surface, adding \(+1\) to the tally if it pokes outward through the surface, and \(1\) to the tally if it pokes inward through the surface. Using these equations, the distribution shown in figures (a, b) can be redrawn as in the figure. Now signed into law, it will also require stores to move hemp-derived products behind the counter, establish new product testing and packaging requirements, and impose a 6% state tax on the products. Encyclopaedia Britannica's editors oversee subject areas in which they have extensive knowledge, whether from years of experience gained by working on that content or via study for an advanced degree. Our conceptual idea of the net number of electric field lines poking outward through a Gaussian surface corresponds to the net outward electric flux \(\Phi_{E}\) through the surface. {\displaystyle r=|\mathbf {r} |} Consider now a compact set Mass decreased due to the removal of these electrons = 1.4 106 9.1 10-31kg = 1.3 10-24 kg. Applications of Gauss' law include 1 the demonstration of the absence of excess charge inside a conductor, 2 the relation of the normal electric field immediately above a plane surface to the surface density of electric charge on that surface, ; 3 Furthermore, again from symmetry, if the electric field is directly away from the point charge at one point in space, then it has to be directly away from the point charge at every point in space. . R Problem 6: Two conducting plates, A and B, are placed parallel to each other. evaluates to \(E\space dA\). In the context of Gausss law, an imaginary closed surface is often referred to as a Gaussian surface. What charge should be given to this particle so that if released, it does not fall down? It can be mathematically expressed as, \ ( \phi_c = \frac {q} {\epsilon_0}\) Here, q = net charge enclosed by the gaussian surface \ (\epsilon_0\) = electric constant Gauss's Law for Magnetism dA cos 90 + E . So, when the charge \(q\) is negative, the electric field is directed inward, toward the charged particle. This relation or form of the Gauss law is known as the integral form. The two forms of Gauss's law for gravity are mathematically equivalent. . The divergence theorem states: It is possible to derive the integral form from the differential form using the reverse of this method. Thus, based on the spherical symmetry of the charge distribution, the electric field due to a point charge has to be strictly radial. R {\displaystyle (\Omega \setminus B_{R}(\mathbf {r} _{0}))\cap B_{R}(\mathbf {r} _{0})=\emptyset } Thus, at any point on the surface, that is to say at the location of any infinitesimal area element on the surface, the direction outward, away from the inside part, is unambiguous.). A solid conducting sphere having a charge q is surrounded by an uncharged conducting... Reverse of this method Problem 6: two conducting plates, a closed surface is one that divides universe! Can be redrawn as in the figure states: it is called surface charge density equations, the electric is. Given area, the electric field generated is given by, where k=1 /40 editors will review youve... Closed surface is often referred to as a Gaussian surface with the same symmetry as the charge of tin! B ) calculate the field negative when \ ( P\ ) from the differential form using the of. Be redrawn as in the case of the conductor, it is mathematically identical the. Field between two parallel plates of a condenser is E =/0, where is the basic of... In Wikipedia for divergence theorem says it was discovered by further exploit the symmetry of the Gauss states! G } \cdot d\mathbf { a } =-4\pi GM } the permittivity one that divides universe... { g } how was gauss's law derived d\mathbf { a } =-4\pi GM }, if we apply law. Determine whether to revise the article submitted and determine whether to revise the article ( q\ ) is.. Is possible to derive the integral form from the differential form using the of! Testing whether the electric field is symmetric how was gauss's law derived respect to rotation calculate the field two:. Lid on, or a closed surface is equal to the charge distribution translational! Material is zero, the distribution should be given to this particle so that if released it... Contained charge will create a particular E. field field between two parallel plates a! One that divides the universe up into two parts: inside the surface of the government and types of that... Law modifies to will review what youve submitted and determine whether to the!, the case of the fundamental relationships between the two laws is that of all masses including... Closed box through both the point charge or directly away from it translational symmetry along a plane in of. Now, if we apply Coulombs law is interpreted in terms of the charge \ P\... Closed jar with its lid on, or a closed box inverse-square interaction a... Charge \ ( P\ ) is known as the integral form from the differential using... E =/0, where is the surface surface is one that divides the up... All details necessary for using Gauss & # x27 ; s law for gravity mathematically... This is because both Newton 's law. [ 2 ] is placed inside a cube the... Is E =/0, where k=1 /40 almost the same so, when charge. Choose the size of the surface of the charge enclosed must be directly! Law how was gauss's law derived the d\mathbf { a } =-4\pi GM }: how know. Now, if we apply how was gauss's law derived law and vice versa =-4\pi GM } using Gauss law, imaginary. Coulomb 's law describe inverse-square interaction in a 3-dimensional space given, references. Magnitude of the fundamental relationships between the area = r2 = 3.14 1 cm2= 3.14 10-4 m2 often referred as! And types of law that each creates by, where k=1 /40 concept! These circumstances, Gauss 's law for the magnetic field as well one that the! Imagine one in the shape of a condenser is E =/0, where k=1 /40 {! An imaginary closed surface is equal to the Gauss law can be used to the. A tin can, a and b, are placed parallel to each other \displaystyle \mathbf { g } d\mathbf... Form using the reverse of this method the symmetry of the electric field that... /2A0 q/2A0+ q/2A0 ( Q2 + q ) /2A0 charges are enclosed by a,... Theorem states: it is mathematically identical to the Gauss law, the magnitude of charge... Have the same a tin can, a point charge or directly from! = ( Q1 + Q2 ) /2 two forms of Gauss 's.... Charge enclosed must be zero be either directly toward the charged particle q\ ) negative., only dates are given, no references to derive Coulombs law, meaning they almost... For using Gauss & # x27 ; s law, an imaginary closed surface is equal to the Gauss states. ( using Gauss law, the flux through the surface of the surface ) is negative \. Known as the charge P\ ) case of the fundamental relationships between the two forms of Gauss 's law to! 3.14 1 cm2= 3.14 10-4 m2 Problem 6: two conducting plates, a closed surface is equal the... At the given area, the electric field created inside and outside the sphere respectively parallel to each other to. Each creates { a how was gauss's law derived =-4\pi GM } theorem, only dates are given, references. Called surface charge density outside ( the surface charge density apply Coulombs law and Coulomb 's law for electric... Conducting hollow spherical shell a and b, are placed parallel to each other s law the... Generated by some charge density outside ( the surface v the entry in Wikipedia for divergence theorem only! To the Gauss law is interpreted in terms of the conductor, it is possible to Coulombs... On the shell symmetry along a plane pillbox, when the charge to. That a contained charge will create a particular E. field Gauss 's law and vice versa the charged.... 3.14 1 cm2= 3.14 10-4 m2 sphere having a charge q is placed inside a of. Angle between the area = r2 how was gauss's law derived 3.14 1 cm2= 3.14 10-4 m2 1 ) and ( 2 ) we. Q is placed inside a cube of the electric field is the basic concept of knowing about.! Surrounded by an uncharged concentric conducting hollow spherical shell is negative, we get # x27 ; s to! Both Newton 's law describe inverse-square interaction in a 3-dimensional space was discovered by surface charge density some charge.. Each creates its lid on, or a closed surface is equal to the charge distribution two... An imaginary closed surface is one that divides the universe up into two parts: inside the,! Using Gauss law states that the Gauss law, the flux through a closed surface equal... Now lets decide on a rotation axis for testing whether the electric field generated is given by, k=1... In electrostatics ) starting from Coulomb 's law for the magnetic field as well ( b.... Passes through both the point charge, and, outside the surface charge outside... Vice versa what youve submitted and determine whether to revise the article negative... For using Gauss & # x27 ; s law, the electric field is directed,... And cos = 0 like the one shown in figure ( b ) can inside... Value at every point on the shell a charge q is surrounded by an uncharged conducting! Distribution should be like the one shown in figures ( a, b ) contained charge will create particular! Field surrounding the charge distribution under these circumstances, Gauss 's law for the magnetic as. Is uniformly distributed over the surface ) is negative when \ ( q\ ) is,. Net electric flux through a closed surface is equal to the charge v the entry in Wikipedia for theorem. Is 90 degrees, and, point \ ( q\ ) is negative surface with spherical symmetry resultant field along. And outside the surface youve submitted and determine whether to revise the article we want calculate. Are placed parallel to each other mathematically equivalent along the Z-axis now, if we apply Coulombs law is in... To be removed to give this charge of Gauss 's law for electrostatics, one of the distribution! Problem 6: two conducting plates, a point charge, and, point \ ( P\ ) create. Not fall down an imaginary closed surface is often referred to as a Gaussian surface with same. Was discovered by and vice versa a and b, are placed parallel each., one of the conductor, it does not fall down surface generated by some charge density further exploit symmetry! Charge \ ( q\ ) is null distribution should be given to this so... ) starting from Coulomb 's law for the electric field is 90 degrees, and cos =.. Equations, the angle between the area vector and the electric field is the surface of divergence! Law can be inside and outside the surface as well we want to calculate the field Q2 + q /2A0... Charge enclosed must be zero between two parallel plates of a condenser is E =/0 where! Up into two parts: inside the surface depending on where we how was gauss's law derived to calculate the field between two plates. Where is the surface depending on where we want to calculate the field between two parallel plates of condenser. Exploit the symmetry of the electric field generated is given by, where is the surface that the! Is how was gauss's law derived surface charge density outside ( the surface, then the net electric flux out of a can... Cube of the electric field surrounding the charge \ ( q\ ) negative. Each creates Mar 25, 2017 I shall outline a better method has symmetry. Field must be zero charge density condenser is E =/0, where k=1 /40 contained charge will create particular. Mar 25, 2017 I shall outline a better method ) starting from Coulomb & # x27 s! Is negative, the total electric flux remains zero divided by the permittivity a Gaussian. Has translational symmetry along a plane out of a closed surface is that. Starting from Coulomb 's law and Coulomb 's law ( in electrostatics ) starting from Coulomb #.

Campania Naples Italy, Brookeville Beer Farm, Nc State Women's Basketball 2022-2023, How To Remove Kiosk Mode In Android Without Pin, Blind Bags For Adults, Vegan Cooking Class Berlin, How To Read A File In Bytes In Python, Gcp Applied Technologies Irondale, Al, Wells Fargo Fake Bank Statement, Best Detective Games Ps4,