Sure, here's how you can use Gauss' law to find the electric field outside of a long straight wire:

1. First, let's assume that the wire is infinitely long and has a uniform linear charge density λ (lambda).

2. Gauss' law states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (ε0). Mathematically, it is represented as ∮E.dA = Q/ε0.

3. To apply Gauss' law, we need to choose a Gaussian surface. In this case, we choose a cylindrical Gaussian surface with the wire as the axis. The cylinder has a length L and radius r.

4. The electric field E due to the wire is directed radially outward from the wire. Therefore, the electric flux through the curved surface of the cylinder is E(2πrL), because E and dA are parallel on this surface.

5. The electric flux through the two end faces of the cylinder is zero, because E and dA are perpendicular on these surfaces.

6. The total charge enclosed by the Gaussian surface is λL, where λ is the linear charge density.

7. According to Gauss' law, the electric flux through the Gaussian surface is equal to the enclosed charge divided by the permittivity of free space. Therefore, we have E(2πrL) = λL/ε0.

8. Solving this equation for E, we get E = λ/(2πε0r).

So, the electric field outside a long straight wire of uniform linear charge density is directed radially outward from the wire and its magnitude decreases with the distance r from the wire as λ/(2πε0r).

Question

Sure, here's how you can use Gauss' law to find the electric field outside of a long straight wire:

1. First, let's assume that the wire is infinitely long and has a uniform linear charge density λ (lambda).

2. Gauss' law states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (ε0). Mathematically, it is represented as ∮E.dA = Q/ε0.

3. To apply Gauss' law, we need to choose a Gaussian surface. In this case, we choose a cylindrical Gaussian surface with the wire as the axis. The cylinder has a length L and radius r.

4. The electric field E due to the wire is directed radially outward from the wire. Therefore, the electric flux through the curved surface of the cylinder is E(2πrL), because E and dA are parallel on this surface.

5. The electric flux through the two end faces of the cylinder is zero, because E and dA are perpendicular on these surfaces.

6. The total charge enclosed by the Gaussian surface is λL, where λ is the linear charge density.

7. According to Gauss' law, the electric flux through the Gaussian surface is equal to the enclosed charge divided by the permittivity of free space. Therefore, we have E(2πrL) = λL/ε0.

8. Solving this equation for E, we get E = λ/(2πε0r).

So, the electric field outside a long straight wire of uniform linear charge density is directed radially outward from the wire and its magnitude decreases with the distance r from the wire as λ/(2πε0r).

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