III.A 3, Gauss’ Law

We have used Coulomb’s Law (the governing law in electrostatics, in case you didn’t know) to derive 𝐸= 𝑘𝑞 𝑟 2 , the electric field of a point charge. However, it the charge is distributed over a plane, cylinder or sphere, we will need another method.

Flux – rate of flow through an area Flux – rate of flow through an area. Could be fluid flow, magnetic field lines or, our current interest, electric field lines. By definition, 𝜙=𝐸𝐴 cos 𝜃 . (N·m2/C)

Four algebraic examples and one calculus example for calculating electric flux.

Ex. Find the net flux through the cylinder in electric field as shown. b a c

Gauss’ Law relates the electric fields at points on a surface (a Gaussian surface) to the net charge enclosed by that surface

Ex. Find the electric flux through a Guassian sphere about a point charge.

So, Gauss’ Law is 𝜙= 𝐸∙𝑑𝐴= 𝑞 𝑒𝑛𝑐 𝜖 𝑜

Gauss Law as shown is only true when the charge is in a vacuum Gauss Law as shown is only true when the charge is in a vacuum. Include the sign of the charge since it gives the direction of the electric flux. If q is positive, the net flux is outward. If q is negative, the net flux is inward.

Ex. In terms of r and linear charge density λ, find the electrical field about a cylinder of uniform charge density. Graph E vs. r for the cylinder.

Ex. In terms of area charge density σ, find the electric field of a thin conducting plate.

Ex. Find the electric field between the two charged conducting plates in terms the area charge density.

Ex. For a uniformly charge conducting spherical shell of inner radius a and outer radius b, find the electric field for a Gaussian surface where r < a, a < r < b, and r > b. Graph E vs. r for the shell.

Ex. Find the electric field for a uniformly charged non-conducting sphere of radius R when r < R and r ≥ R. Graph E vs. r for the sphere.

Ex. A non-conducting sphere or radius R has a non-uniform charged density described by the function 𝝆 𝒓 = 𝝆 𝒐 𝒓 𝒂 𝟐 where ρO and a are constants. Find the electric field when r < R and r ≥ R. Graph E vs. r for the sphere.

Charged isolated conductor If excess charge is placed on an isolated conductor, all the excess charge will move to the surface of the conductor. Therefore, E inside this conductor is zero. Conductor with a cavity. Excess charge resides on the surface and E inside the conductor is zero. A shell of uniform charge attracts or repels a charged particle outside the sphere as if the charge were concentrated at the center of the sphere.

 

Ex. Use Gauss’ Law to show the electric field inside a charged hollow metal sphere is zero.

Ex. A solid metal sphere with radius a, carrying a charge of +q is placed inside, and concentric with a neutral hollow metal sphere of inner radius b and outer radius c. Determine the electric field for r < a, a < r < b, b < r < c, and r > c.

Parallel Plate Capacitors

A derivation

So, you (maybe) can see capacitance measures the capacity for: holding charge; storing electrical potential energy; or storing an electric field. Capacitance depends upon the geometry of the capacitor

Ex. A 10-nF parallel-plate capacitor holds a charge of magnitude 50 μC on each plate. a) What is the potential difference between the plates? b) If the plates are separated by a distance of 0.2 mm, what is the area of each plate?

Capacitors of other geometries

Ex. A long cable consists of a solid conducting cylinder of radius a, which carries a linear charge density of +λ, concentric with an outer cylindrical shell with inner radius b, which carries a charge density –λ. This is a coaxial cable. Determine the capacitance of the cable.

Ex. A spherical conducting shell of radius a, which carries charge +Q, is concentric with an out spherical conducting shell of inner radius b and carries a charge of –Q. What is the capacitance of the spherical capacitor?