Electric field in the cylindrical capacitor using Gauss' Law [Cylindrical Shell Capacitor Pt. 1]
In this video, we compute the electric field in the cylindrical capacitor using Gauss' Law (also called the "coaxial capacitor").
This video is Part 1 of a three-part series on the cylindrical shell capacitor:
Cylindrical Shell Capacitor Pt. 1: electric field in the cylindrical shell capacitor.
Cylindrical Shell Capacitor Pt. 2: capacitance of the cylindrical shell capacitor: • Capacitance per unit length in the cy...
Cylindrical Shell Capacitor Pt. 3: energy stored in the cylindrical shell capacitor: • Energy in the cylindrical capacitor i...
We are given a long cylindrical shell capacitor, consisting of two concentric cylindrical conducting shells. The inner and outer radii are given, and we are given the linear charge density on the inner shell as +lambda and the linear charge density on the outer shell as -lambda. The assumption that the cylindrical shells are long is what guarantees the electric field has good symmetry, pointing radially outward from the inner shell to the outer shell.
To use Gauss' Law with cylindrical symmetry we choose a concentric cylinder as a gaussian surface. We comment that a gaussian cylinder with r less than a would enclose zero charge, so the electric field inside the inner shell must be zero. Similarly, a gaussian cylinder with r greater than b encloses zero charge because there are equal parts positive and negative charge enclosed (zero net charge). The region of interest is in between the shells, so we show a gaussian surface with a radius between a and b.
Next, we invoke Gauss' Law, and we give a link to the derivation of Gauss' Law if you would like a quick refresher: • Electric flux, derivation of Gauss' L...
We compute the electric flux integral, noting that the end-caps of the gaussian surface give us zero flux because the electric field is perpendicular to the area vector. All the flux happens through the curved surface of the cylinder. Along this surface, E has a constant magnitude and is always parallel to the area vector, so the dot product is trivial and E can be factored out, leaving us with E*(2pi*r)L for the total electric flux.
The enclosed charge is lambda*L, and we solve for E, arriving that the electric field inside the cylindrical capacitor: E=lambda/(2pi*epsilon_0*r).
