Electrostatics Review Sheet for Midterm 1 - Electric Field and Potential, Study notes of Electromagnetism and Electromagnetic Fields Theory

Download Electrostatics Review Sheet for Midterm 1 - Electric Field and Potential and more Study notes Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! 1 PHY481 - Review sheet for Midterm 1 Griffiths: Chapters 1-3 Electric Field There are two types of charge and they interact through Coulomb’s law ~F = 14π0 qQ r2 r̂ = 1 4π0 qQ r3 ~r. The interaction between many charges is found by using superposition. The electric field due to a charge Q through, ~E = ~Fq , so that the electric field is the force per unit charge. Since the unit of force is the Newton (N), the unit of electric field is N/C, where C is the unit of electric charge, the Coulomb. The electric field at position ~r due to a point charge at position ~r′ is ~E(~r, ~r′) = k Q|~r−~r′|3 (~r − ~r ′) The superposition principle indicates that the electric field at position ~r due to n charges at positions ~r1...~rn is given by the vector sum, ~E(~r) = n∑ i=1 k Qi |~r − ~ri|3 (~r − ~ri) = k ∫ d~r′ρ(~r′) |~r − ~r′|3 (~r − ~r′) (1) where ρ(~r′) = ∑ iQiδ(~r ′ − ~ri). When treating continuous charge distributions, we may be given a charge per unit length, λ, a charge per unit area σ or a charge per unit volume ρ. Typical superposition problems: ring of charge; disc of charge; line segment etc. Note that you can always solve the superposition problem for the potential and then take a gradient to find the electric field. This is often easier. An electric field line is a series of vectors where at each point the vector points in the direction of the force on a unit charge at that point and it has a length equal to the magnitude of the force. ie. we plot the vector function ~E. The properties of electric field lines constructed in this way are as follows. (i) At each point along an electric field line, the force on a positive test charge is in a direction tangent to the field line at that point. This implies that electric field lines come out of positive charges and go into negative charges. (ii) The density of lines at any point in space is proportional to the magnitude of the electric field at that point. (iii) Electric field lines begin and/or end at charges, or they continue off to infinity. i.e. they do not begin or end in free space. (iv) Electric field lines do not cross. Conductors: If there is no current flowing, then the electric field is zero, ~E = 0, inside a conductor, and at the surface of a conductor the electric field is normal to the surface (know the reasoning behind this). The integral form of Gauss’s law in free space is, φE = ∮ S ~E · d~a = qencl 0 (2) where qencl = ∑ iετ qi = ∫ τ ρ(~r)d~r where ρ(~r) is the charge density. The differential form of Gauss’s law: ~∇· ~E = ρ/0 (know how to derive this). Know how to solve Gauss’s law problems in spherical, cylindrical and planar geometries, including at a conducting surface. It is good memorize some of these: kQr̂/r2; λŝ/(2π0s); σn̂/(20); σn̂/0 and their derivations. Another useful result is the electric field inside a uniform sphere of charge ρ~r/(30). Electrostatic potential The difference in potential energy between two positions a and b is ∆Uab = ∫ b a ~Fext · ~dl = − ∫ b a ~F · ~dl. We define Vab = Uab/q so that ∆Vab = − ∫ b a ~E · ~dl, (3) for any path between a and b. This definition is consistent with the differential form, where ~∇ ∧ ~E = 0 implies that ~E = −~∇V . Since ~∇ · ~E = ρ/0, we have ∇2V = −ρ/0 (Poisson’s equation). The special case ρ = 0 is Laplace’s equation. The superposition formula for the potential is V (~r) = ∑ i kqi |~r − ~ri| → ∫ d~r′ kρ(~r′) |~r − ~r′| . (4) Typical superposition problems: ring of charge; disc of charge; line of charge; shell of charge; uniform sphere of charge. It is good to memorize some of the results: kQ/r (outside a uniform sphere or shell of charge); −λln(s)/(2π0) + C (outside a line or uniform cylinder of charge); −E0z+C (sheet of charge or uniform electric field in z-direction). You should know how to derive these by integration of the Gauss’s law results. The potential energy of a small element of charge in a potential is U = qV → ∫ ρ(~r)V (~r) (constant V ), however the energy required to set up a charge distribution is Un = 12 ∑n,n i 6=j kQiQj rij = ∑n,n i<j kQiQj rij → 12 ∫ ρ(~r)V (~r). Each pair