INTRO:**Gauss”s law relates the electric flux ΦEthrough a closed surface to the total charge qenclenclosed by the surface:ΦE=∮E⋅dA = qencl/ϵ0.You can use Gauss”s law to determine the charge enclosed inside a closed surface on which the electric field is known. However, Gauss”s law is most frequently used to determine the electric field from a symmetric charge distribution.The simplest case in which Gauss”s law can be used to determine the electric field is that in which the charge is localized at a point, a line, or a plane. When the charge is localized at a point, so that the electric field radiates in three-dimensional space, the Gaussian surface is a sphere, and computations can be done in spherical coordinates. Now consider extending all elements of the problem (charge, Gaussian surface, boundary conditions) infinitely along some direction, say along thezaxis. In this case, the point has been extended to a line, namely, thezaxis, and the resulting electric field has cylindrical symmetry. Consequently, the problem reduces to two dimensions, since the field varies only withxandy, or with r and θ in cylindrical coordinates. A one-dimensional problem may be achieved by extending the problem uniformly in two directions. In this case, the point is extended to a plane, and consequently, it has planar symmetry. **

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**—————————————————————————————————–Three dimensionsConsider a point charge q in three-dimensional space. Symmetry requires the electric field to point directly away from the charge in all directions. To find E(r), the magnitude of the field at distance r from the charge, the logical Gaussian surface is a sphere centered at the charge. The electric field is normal to this surface, so the dot product of the electric field and an infinitesimal surface element involves cos(0)=1. The flux integral is therefore reduced to ∫E(r)⋅dA=E(r)⋅A(r), where E(r) is the magnitude of the electric field on the Gaussian surface, and A(r) is the area of the surface. (Figure 1) PART A:Determine the magnitude E(r) by applying Gauss”s law.Express E(r) in terms of some or all of the variables/constants q, r, and ϵ0.** SOLUTION:**E(r) = q/(4πr2ε0)HINTS:**

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**Find the area of a spherical surface of radius r surrounding the point charge.Express your answer in terms of r and other constants.A(r) = 4πr2 —————————————————————————————————–Two dimensionsNow consider the case that the charge has been extended along thezaxis. This is generally called a line charge. The usual variable for a line charge density (charge per unit length) is λ, and it has units (in the SI system) of coulombs per meter. PART B:By symmetry, the electric field must point radially outward from the wire at each point; that is, the field lines lie in planes perpendicular to the wire. In solving for the magnitude of the radial electric field E(r) produced by a line charge with charge density λ, one should use a cylindrical Gaussian surface whose axis is the line charge. The length of the cylindrical surface L should cancel out of the expression for E(r). Apply Gauss”s law to this situation to find an expression for E(r). (Figure 2)Express E(r) in terms of some or all of the variables λ, r, and any needed constants.**** SOLUTION:E(r) = λ/(2πrε0)HINTS:Find A(r), the area of the Gaussian surface. Note that you do not need to include the ends of the cylinder in the calculation of area. Since the electric field is radial (by symmetry), the ends of the cylinder are parallel to the field, and there is no flux through them.Express A(r) in terms of the length L and radius r of the surface, as well as any needed constants.A(r) = 2πrL**

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**What is the charge qenclcontained within the Gaussian cylinder?Express your answer in terms of λ, L, and any needed constants.qencl= λL—————————————————————————————————–One dimensionNow consider the case with one effective direction. In order to make a problem effectively one-dimensional, it is necessary to extend a charge to infinity along two orthogonal axes, conventionally taken to bexandy. When the charge is extended to infinity in thexyplane (so that by symmetry, the electric field will be directed in thezdirection and depend only onz), the charge distribution is sometimes called a sheet charge. The symbol usually used for two-dimensional charge density is either σ, or η. In this problem we will use σ. σ has units of coulombs per meter squared. PART C:In solving for the magnitude of the electric field E(z) produced by a sheet charge with charge density σ, use theplanarsymmetry since the charge distribution doesn”t change if you slide it in any direction ofxyplane parallel to the sheet. Therefore at each point, the electric field is perpendicular to the sheet and must have the same magnitude at any given distance on either side of the sheet. To take advantage of these symmetry properties, use a Gaussian surface in the shape of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A which will cancel out of the expression for E(z) in the end. The result of applying Gauss”s law to this situation then gives an expression for E(z) for both z>0 and z Express E(z) for z>0 in terms of some or all of the variables/constants σ, z, and ϵ0.** SOLUTION:E(z) = σ/(2ε0) NOTES:In this problem, the electric field from a distribution of charge in 3, 2, and 1 dimension has been found using Gauss”s law. The most noteworthy feature of the three solutions is that in each case, there is a different relation of the field strength to the distance from the source of charge. In each case, the field strength varies inversely as an integral power of the distance r from the charge. In the case of a point charge (spherical symmetry, field in three dimensions), the field strength varies as r−2. In the case of a line charge (cylindrical symmetry, field in two dimensions), the field strength varies as r−1. Finally, in the case of a sheet charge (planar symmetry, field in one dimension), the field varies as r0=1; that is, the strength of the field is independent of the distance from the sheet!If you visualize the electric field using field lines, this result shows that as the number of directions in which the electric field can point is reduced, the field lines have one dimension fewer in which to spread out, and the field, therefore, falls off less rapidly with distance. In a one-dimensional problem (sheet charge), the extension of the charge in thexy-plane means that all field lines are parallel to thez-axis, and so the field strength does not change with distance. Such a situation, of course, is impossible in the real world: In reality, the planar charge is not infinite, so the field will, in fact, fall off over long distances.HINTS:

What is the total flux ΦEout of the ends of the cylinder used as the Gaussian surface in this problem? Note that since the electric field is directed along thezaxis by symmetry, it is parallel to the curvedsidewalls of the cylinder, so there is no flux through these walls.Express your answer in terms of the area A of the ends of the cylinder, the magnitude E of the electric field, and any needed constants.ΦE= 2EA

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What is the charge qenclinside the Gaussian surface?Express your answer in terms of σ, A, and any needed constants.qencl= σA