Question

Gauss's Law in 3, 2, and 1 Dimension 

Gauss's law relates the electric flux \(\Phi_{E}\) through a closed surface to the total charge \(q_{\text {end }}\) enclosed by the surface:

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 \(q, \tau\), and \(\epsilon_{0}\).

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 \(\lambda\), 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 \(\lambda, r\), and any needed constants.

Part C

In solving for the magnitude of the electric field \(\vec{E}(z)\) produced by a sheet charge with charge density \(\sigma\), use the planar symmetry since the charge distribution doesn't change if you slide it in any direction of \(x y\) 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 \(\sigma, z\), and \(\epsilon a\)

Answer 1

Hope this helps..