Question

6. Using Gauss’s Law, find the electric field a distance r from a line of positive charge of infinite length and a constant charge per unit length λ. Draw a diagram, a Gaussian surface and completely prove your answer

Answer 1

let us determine the electric field intensity E at a distance r from an infinite line of charge lying on the x-axis. A small portion of the infinite line of charge is shown in figure (a).

The charge is uniformly distributed and has a linear charge density λ. (Recall that the linear charge density λ is the charge per unit length, that is, λ = q/L.) To determine the electric field we must first draw a Gaussian surface. The line of charge does have a cylindrical symmetry however. Therefore, we draw a cylindrical Gaussian
surface of radius r and length L around the infinite line of charge as shown in figure (b). Gauss’s law for the total flux emerging from the Gaussian cylinder,

   _____________ (1)

The integral in Gauss’s law is over the entire Gaussian surface. We can break the entire cylindrical Gaussian surface into three surfaces. Surface I is the end cap on the left-hand side of the cylinder, surface II is the main cylindrical surface, and
surface III is the end cap on the right-hand side of the cylinder as shown in figure (b). The total flux Φ through the entire Gaussian surface is the sum of the flux through each individual surface. That is,
Φ = Φ_I + Φ_II + Φ_III _________________ (2)
where
Φ_I is the electric flux through surface I
Φ_II is the electric flux through surface II
Φ_III is the electric flux through surface III
Hence, Gauss’s law becomes

______ (3)  

Notice that along cylindrical surface I, is everywhere perpendicular to the surface and points toward the left as shown in figure (b). The electric field intensity vector lies in the plane of the end cylinder cap and is everywhere perpendicular to the surface vector dA of surface I and hence θ = 90. Therefore the electric flux through surface I is
______ (4)
Surface II is the cylindrical surface itself. As can be seen in figure (b), E is everywhere perpendicular to the cylindrical surface pointing outward, and the area vector dA is also perpendicular to the surface and also points outward. Hence E and dA are parallel to each other and the angle θ between and is zero. Hence, the flux through surface II is

   ______ (5)

Along cylindrical surface III, is everywhere perpendicular to the surface and points toward the right as shown in figure (b). The electric field intensity vector E lies in the plane of the end cylinder cap and is everywhere perpendicular to the surface vector dA of surface III and therefore θ = 90. Hence the electric flux through surface III is

   ________ (6)

Combining the flux through each portion of the cylindrical surfaces using (4) ,(5) & (6) , equation (3) becomes

   _________ (7)

From the symmetry of the problem, the magnitude of the electric field intensity E is a constant for a fixed distance r from the line of charge, that is, there is no reason to assume that E in the z-direction is any different than E in the y-direction. Hence, E can be taken outside the integral sign to yield
  
The represents the sum of all the elements of area dA, and that sum is just equal to the total area of the cylindrical surface. It is easier to see the total area,if we unfold the cylindrical surface as shown in figure (c). One length of the
surface is L, the length of the cylinder, while the other length is the unfolded circumference 2πr of the end of the cylindrical surface.The total area A is just the product of the length times the width of the rectangle formed by unfolding the
cylindrical surface, that is, A = (L)(2πr). Hence, the integral of dA is = A = (L)(2πr)
Thus, Gauss’s law becomes
  
Solving for the magnitude of the electric field E intensity we get

But q/L = λ the linear charge density and using the fact that Electric Field Vector is always in the radially outward direction () Therefore