Gauss s Law Johann Carl Friedrich Gauss 1777-1855 Mathematics and physics: including number theory, analysis, differential geometry, geodesy, electricity, magnetism, astronomy and optics. y = Be cx 2
First, electric flux, Φ E Flux means flow. In this case we are interested in number of field lines passing through a given surface. More precisely, flux is found by taking the component of the electric field perpendicular to a given surface, and multiplying by the area of the surface. (Or, equivalently, the surface area perpendicular to the electric field, times the field, E.) Φ 0 o r r 90 EA Φ = E A = EA E( Acosφ) o Φ E = 0 E = E = General expression for a uniform field passing through a flat surface
Dimensions of electric flux, E Φ [ ] [ ] [ ] [ ] m V m m V C m N m C N A m C N E E = = = = Φ 2 2 2 2 : Dimensions
Example: electric flux,, for a uniform field through a disk Φ E E = 500 N/C Do
The general expression for Φ E Unless the surface is flat and the field is uniform over the surface, we must calculate the flux by integrating the perpendicular (normal) component of the electric field over the surface. Φ E = A v r E da = E cos(φ ) da A But most of the problems we do will involve simple surfaces and uniform or symmetric electric fields!
Gauss s Law: here it is! The electric flux through a closed surface is proportional to the total charge in the enclosed volume. Wunderbar! Φ E = Q enclosed ε 0 Constant of proportionality = 1/ε 0 Sometimes the closed surface is that of a real physical object. Sometimes we create a closed surface purely for doing the calculation: a Gaussian surface. Onward, to the examples
The simplest case: a single point charge We have chosen to surround the point charge with a spherical Gaussian surface of radius r centered on the charge. This means that the electric field is perpendicular to the surface, and is the same magnitude at all points on the surface. We need not integrate. Φ EA Q enclosed E = = = q ε 0 ε 0 q r Since A = 4π r 2 q 2 EA = E(4πr ) = E = = 2 ε 0 4πr ε 0 q kq 2 r This is the electric field for a point charge--found from Coulomb s Law!!
Gauss s Law, with point charges in boxes In these cases, it is easy to find Q enclosed simply by adding point charges. But calculating the total flux through the closed surface of a box is complicated since E changes from point to point. We must integrate without help from symmetry: Φ E = A v r E da = E cos(φ ) da A Then we could apply Gauss s Law Φ E = Q enclosed ε 0
Gauss s Law, more charges in boxes There are a number of ways to get zero flux. Independent of angle If the charge in the box is constant, the total flux through the surface is constant, even if the surface is changed.
Gauss s Law, more general rules & examples Flux near point charges Zero net flux, generally. Gaussian surfaces near a dipole
Choosing a Gaussian surface wisely With this choice of Gaussian surface, E is constant in magnitude, and perpendicular to the surface, at all points. (We did the calculation.) With this choice of Gaussian surface, we would calculate the same total flux, but with E changing in magnitude and direction from point to point, we would need to do a full-blown surface integral: v r Φ E = A E da = E cos(φ ) da A
Electric field of a sheet of charge of uniform charge density, sigma We choose the surface so that some faces have zero flux through them, and some have perpendicular electric fields making the flux calculation easy. σ Do
Electric field of a charged conducting plate: two sheets of charge Slightly different from the previous problem! Do
Do Electric fields of a capacitor: two conducting plates
Electric field of a line charge of uniform density, λ We again choose the surface so that some faces have zero flux through them, and some have perpendicular electric fields. Do
Do Electric field of a solid (insulating) sphere of uniform charge density, ρ
Electric field of a solid sphere (above)
Conductors in electrostatic equilibrium: electric field properties from self-consistency I. In electrostatic equilibrium, there is no field inside a conductor. Otherwise, the charges inside would continue to move! (F = qe) II. Starting point: In conductors, charges are free to move, both inside and on the surface. When they have stopped moving, the excess charges are in electrostatic equilibrium. In electrostatic equilibrium, all excess charges are on the surface of a conductor. Otherwise, there would be electric fields inside! III. In electrostatic equilibrium, the electric field outside the conductor, and near the surface, is perpendicular to the surface. Otherwise, excess charges would continue to move along the surface! Sketch an oddly shaped conductor being charged, and then in equilibrium.
Conductors in electrostatic equilibrium: electric field properties and Gauss s Law Gauss s Law is consistent with properties I. and II. above.
Conductors in electrostatic equilibrium: electric field properties and Gauss s Law We can use Gauss s Law, together with property III., to find the electric field at any point near the surface of a conductor, independent of shape σ
Conductor examples: using Gauss s Law
Faraday s Pail, or Cup
Faraday Cage
We ll see more of Faraday s work as we go along. From Wikipedia: Michael Faraday, (1791 1867) was an English chemist and physicist who contributed significantly to the fields of electromagnetism and electrochemistry. He established that magnetism could affect rays of light and that there was an underlying relationship between the two phenomena. Some historians of science refer to him as the best experimentalist in the history of science. It was largely due to his efforts that electricity became viable for use in technology. The SI unit of capacitance, the farad, is named after him. Faraday's law of induction states that a magnetic field changing in time creates a proportional electromotive force.
When lightning strikes Nature, discharging its capacitors. Faraday s Car (Not really. It just looks that old.) 1950 Ford sedan. Discuss.
Calculating the surface charge at each point on the surface of a conductor We did this calculation before, when considering the field at the surface of a flat conducting plate. The same equation, E = σ/ε 0, applies here. But σ and E now vary from one point to the next on the surface.
The Van de Graaf generator
Spherical shell with point charge at center Find and sketch the field for (a) a conducting shell, and (b) an insulating shell Do
Comparing electric fields of insulators and conductors: example of spherical shells charged to +Q and -Q +Q R r R -Q d 1. Consider insulators at distances: d ~ R, d >> R, and d << R. 2. Consider conductors at distances: d ~ R, d >> R, and d << R. Do
Do Insulating sphere of uniform charge density, with spherical cavity
Discuss Inverse square law again with Gaussian surfaces
Gauss s Law with both sides in integral form Φ E = enclosed E da = = A r r Q 1 ε ε 0 0 V ρdv In this form, it is clear that Gauss s Law can be used to analyze any shape of closed surface enclosing any distribution of charge. In mathematical terms, this equation is a special case of the Divergence Theorem, also known as Gauss s Theorem. The Divergence Theorem is one of a several integral equations from the area of vector calculus.