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  4. Gauss's law
  5. Applying gauss’s law

6.3 Applying gauss’s law  (Page 3/13)

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Figure shows a cylinder of length L. A line perpendicular to the axis connects the axis to point P on the surface of the cylinder. An arrow labeled delta vector A points outward from P in the same direction as the line. Another arrow labeled vector E subscript P originates from the tip of the first arrow and points in the same direction. A third arrow labeled delta vector A points outward from the top surface of the cylinder, perpendicular to it. An arrow labeled vector E originates from the base of the third arrow and is perpendicular to it.
The Gaussian surface in the case of cylindrical symmetry. The electric field at a patch is either parallel or perpendicular to the normal to the patch of the Gaussian surface.

The electric field is perpendicular to the cylindrical side and parallel to the planar end caps of the surface. The flux through the cylindrical part is

∫ S E → · n ^ d A = E ∫ S d A = E ( 2 π r L ) ,

whereas the flux through the end caps is zero because E → · n ^ = 0 there. Thus, the flux is

∫ S E → · n ^ d A = E ( 2 π r L ) + 0 + 0 = 2 π r L E .

Using gauss’s law

According to Gauss’s law, the flux must equal the amount of charge within the volume enclosed by this surface, divided by the permittivity of free space. When you do the calculation for a cylinder of length L , you find that q enc of Gauss’s law is directly proportional to L . Let us write it as charge per unit length ( λ enc ) times length L :

q enc = λ enc L .

Hence, Gauss’s law for any cylindrically symmetrical charge distribution yields the following magnitude of the electric field a distance s away from the axis:

Magnitude: E ( r ) = λ enc 2 π ε 0 1 r .

The charge per unit length λ enc depends on whether the field point is inside or outside the cylinder of charge distribution, just as we have seen for the spherical distribution.

Computing enclosed charge

Let R be the radius of the cylinder within which charges are distributed in a cylindrically symmetrical way. Let the field point P be at a distance s from the axis. (The side of the Gaussian surface includes the field point P .) When r > R (that is, when P is outside the charge distribution), the Gaussian surface includes all the charge in the cylinder of radius R and length L . When r < R ( P is located inside the charge distribution), then only the charge within a cylinder of radius s and length L is enclosed by the Gaussian surface:

λ enc L = { ( total charge ) if r ≥ R ( only charge within r < R ) if r < R .

Uniformly charged cylindrical shell

A very long non-conducting cylindrical shell of radius R has a uniform surface charge density σ 0 . Find the electric field (a) at a point outside the shell and (b) at a point inside the shell.

Strategy

Apply the Gauss’s law strategy given earlier, where we treat the cases inside and outside the shell separately.

Solution

  1. Electric field at a point outside the shell. For a point outside the cylindrical shell, the Gaussian surface is the surface of a cylinder of radius r > R and length L , as shown in [link] . The charge enclosed by the Gaussian cylinder is equal to the charge on the cylindrical shell of length L . Therefore, λ enc is given by
    λ enc = σ 0 2 π R L L = 2 π R σ 0 .

    Two cylinders sharing the same axis are shown. The outer one has length L, which is smaller than the inner cylinder’s length. A line perpendicular to the axis connects the axis to point P on the surface of the outer cylinder. An arrow labeled r hat points outward from P in the same direction as the line. Another arrow labeled vector E subscript out originates from the tip of the first arrow and points in the same direction.
    A Gaussian surface surrounding a cylindrical shell.

    Hence, the electric field at a point P outside the shell at a distance r away from the axis is
    E → = 2 π R σ 0 2 π ε o 1 r r ^ = R σ 0 ε o 1 r r ^ ( r > R )

    where r ^ is a unit vector, perpendicular to the axis and pointing away from it, as shown in the figure. The electric field at P points in the direction of r ^ given in [link] if σ 0 > 0 and in the opposite direction to r ^ if σ 0 < 0 .
  2. Electric field at a point inside the shell. For a point inside the cylindrical shell, the Gaussian surface is a cylinder whose radius r is less than R ( [link] ). This means no charges are included inside the Gaussian surface:
    λ enc = 0 .

    Two cylinders sharing the same axis are shown. The inner one has length L, which is smaller than the outer cylinder’s length. An arrow labeled E subscript in originates from a point P on the inner cylinder and points outward, perpendicular to the axis.
    A Gaussian surface within a cylindrical shell.

    This gives the following equation for the magnitude of the electric field E in at a point whose r is less than R of the shell of charges.
    E in 2 π r L = 0 ( r < R ) ,

    This gives us
    E in = 0 ( r < R ) .

Significance

Notice that the result inside the shell is exactly what we should expect: No enclosed charge means zero electric field. Outside the shell, the result becomes identical to a wire with uniform charge R σ 0 .

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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