Obtain an expression for the intensity of electric field due to an inf...
Electric field intensity due to an infinitely long, straight uniformly charged wire using Gauss's law
Gauss's law states that the electric flux through any closed surface is directly proportional to the charge enclosed within the surface. We can use this law to derive an expression for the electric field intensity due to an infinitely long, straight uniformly charged wire at a point near it.
Setting up the problem
Consider an infinitely long, straight wire with uniform charge density ρ. We want to find the electric field intensity at a point P located at a distance r from the wire.
Choosing a Gaussian surface
We choose a cylindrical Gaussian surface with radius r and length l, centered at the wire and passing through the point P. The surface area of this cylinder is given by:
A = 2πrl
Calculating the charge enclosed
The charge enclosed within the Gaussian surface is given by:
Qenc = ρAl
where A is the cross-sectional area of the wire and l is the length of the Gaussian surface.
Calculating the electric flux
The electric flux through the Gaussian surface is given by:
ΦE = E.A = EA
where E is the electric field intensity at point P and A is the surface area of the Gaussian surface.
Applying Gauss's law
According to Gauss's law, the electric flux through the Gaussian surface is equal to the charge enclosed within the surface divided by the permittivity of free space:
ΦE = Qenc / ε0
Substituting the values of ΦE and Qenc from the previous equations, we get:
EA = ρAl / ε0
Deriving the expression for the electric field intensity
Solving for E, we get:
E = ρ / (2πε0r)
This is the expression for the electric field intensity due to an infinitely long, straight uniformly charged wire at a point near it.
Conclusion
We can use Gauss's law to derive an expression for the electric field intensity due to an infinitely long, straight uniformly charged wire at a point near it. The expression is given by E = ρ / (2πε0r).