Gauss's law relates the electric flux ΦE through a closed surface to the total charge qenclenclosed by the surface:
ΦE=∮E⃗ ⋅dA⃗ =qenclϵ0.

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 the z axis. In this case, the point has been extended to a line, namely, the z axis, and the resulting electric field has cylindrical symmetry. Consequently, the problem reduces to two dimensions, since the field varies only with x and y, 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.

Three dimensions

Consider 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.

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E(r) =
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Two dimensions

Now consider the case that the charge has been extended along the z axis. 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.

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E(r) =
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One dimension

Now 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 be x and y. When the charge is extended to infinity in the xy plane (so that by symmetry, the electric field will be directed in the z direction and depend only on z), 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 E⃗ (z) produced by a sheet charge with charge density σ, use the planar symmetry since the charge distribution doesn't change if you slide it in any direction of xy plane 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<0. (Figure 3)

Express E(z) for z>0 in terms of some or all of the variables/constants σ, z, and ϵ0.

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E(z) =