How do you find the electric field intensity of an infinite wire?

You can find the electric field intensity (E) of an infinite wire using Gauss's Law, which relates the electric flux through a closed surface to the enclosed charge.

Derivation using Gauss's Law

Here's the step-by-step derivation:

1. Assume an Infinitely Long Wire: Imagine an infinitely long, straight wire with a uniform linear charge density λ (charge per unit length).

2. Choose a Gaussian Surface: Due to the symmetry of the problem (cylindrical symmetry), a cylindrical Gaussian surface is most convenient. This cylinder has radius 'r' and length 'L', coaxial with the wire.

3. Apply Gauss's Law: Gauss's Law states:

   ∮ E ⋅ dA = Q_enclosed / ε₀

   where:

   • E is the electric field vector
   • dA is the differential area vector (normal to the surface)
   • Q_enclosed is the charge enclosed within the Gaussian surface
   • ε₀ is the permittivity of free space

4. Evaluate the Surface Integral: The cylindrical Gaussian surface has three parts: the top, the bottom, and the curved side. Let's consider each:

   • Top and Bottom: The electric field E is radial and perpendicular to the axis of the wire. Therefore, E is parallel to the curved surface but perpendicular to the top and bottom surfaces. Thus, E ⋅ dA = 0 on the top and bottom surfaces, and the integral over these surfaces is zero.

   • Curved Side: On the curved side, E is parallel to dA. Also, the magnitude of E is constant at a given distance 'r' from the wire. Therefore, E ⋅ dA = E dA. The integral over the curved surface becomes:

     ∮ E ⋅ dA = E ∮ dA = E (2πrL)

     where 2πrL is the surface area of the curved side of the cylinder.

5. Calculate the Enclosed Charge: The charge enclosed within the Gaussian surface (a cylinder of length L) is:

   Q_enclosed = λL

6. Apply Gauss's Law Equation: Substitute the results from steps 4 and 5 into Gauss's Law:

   E (2πrL) = λL / ε₀

7. Solve for E: Solve for the electric field intensity E:

   E = λ / (2π ε₀ r)

8. Vector Form: Expressing this in vector form, with r̂ as the unit vector pointing radially outward from the wire:

   E = (λ / (2π ε₀ r)) r̂

Final Answer

Therefore, the electric field intensity (E) of an infinitely long wire with uniform linear charge density λ at a distance 'r' from the wire is:

E = λ / (2π ε₀ r) or E = (λ / (2π ε₀ r)) r̂

Which can also be represented as:

*E = 1 / (2 π ϵ₀) λ / r**

Or

*E = 1 / (4 π ϵ₀) 2λ / r**

Where:

• E is the electric field intensity.
• λ is the linear charge density (charge per unit length) of the wire.
• r is the distance from the wire to the point where the electric field is being calculated.
• ε₀ is the permittivity of free space (approximately 8.854 x 10^-12 C²/Nm²).
• r̂ is the unit vector pointing radially outward from the wire.