23.6:
Gauss’s Law: Cylindrical Symmetry
Consider different infinitely long straight cylinders, each having distinct charge distributions.
Among these, only the systems with a charge density that does not vary when you rotate it and does not vary along the axis length possess cylindrical symmetry. In comparison, the others do not have cylindrical symmetry.
When there is a cylindrically symmetric charge distribution, a cylindrical Gaussian surface is constructed to obtain the electric flux.
The electric field through the curved part of this surface is parallel to the area vector and has the same magnitude over the circumference and length. From this, the flux over the curved portion is obtained.
The electric field through the flat ends is perpendicular to the area vector, making the flux zero. By combining both, the total flux through the Gaussian surface is obtained.
Since the charge density is constant over the Gaussian cylinder length, the charge enclosed is the product of line charge density and cylinder length.
According to Gauss' law, the electric field magnitude is obtained, which varies inversely with distance from the line charge.
23.6:
Gauss’s Law: Cylindrical Symmetry
A charge distribution has cylindrical symmetry if the charge density depends only upon the distance from the axis of the cylinder and does not vary along the axis or with the direction about the axis. In other words, if a system varies if it is rotated around the axis or shifted along the axis, it does not have cylindrical symmetry. In real systems, we do not have infinite cylinders; however, if the cylindrical object is considerably longer than the radius from it that we are interested in, then the approximation of an infinite cylinder becomes useful.
In all cylindrically symmetrical cases, the electric field at any point P must also display cylindrical symmetry. To make use of the direction and functional dependence of the electric field, a closed Gaussian surface in the shape of a cylinder with the same axis as the axis of the charge distribution is chosen. The flux through this surface of radius r and height L is easy to compute if we divide our task into two parts: (a) the flux through the flat ends and (b) the flux through the curved surface. The electric field is perpendicular to the cylindrical side and parallel to the planar end caps of the surface. The flux is only due to the cylindrical part whereas the flux through the end caps is zero because the area vector is perpendicular to the electric field. Thus, the flux is
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. For a cylinder of length L, the charge enclosed by the cylinder is the product of the charge per unit length and the cylinder length. Hence, Gauss’s law for any cylindrically symmetrical charge distribution yields the following magnitude of the electric field at a distance r away from the axis:
Suggested Reading
- OpenStax. (2019). University Physics Vol. 2. [Web version]. Retrieved from 6.4: Applying Gauss’s Law – Physics LibreTexts.