Download Electrostatics: Coulomb's Law and Electric Field - Prof. Phillip Duxbury and more Study notes Physics in PDF only on Docsity! PHY481 - Lecture 5 Sections 3.1-3.5 of PS All of electrostatics follows from Coulomb’s law + superposition A. Coulomb’s law - Force between two charges The starting point in electrostatics is Coulomb’s law, which gives the force between two stationary charges, ~F = k Qq r2 r̂ = k Qq r3 ~r (1) - Q, q are stationary charges. Their units are coulombs (C) - r̂ is a unit vector along the line between the two charges. - ~r is the vector distance between the two charges. - k = 9 × 109Nm2/C2 = 9 × 109kgm3/C2s2. - k = 1/4πǫ0. ǫ0 is the permittivity of free space. B. Force between many charges - superposition Force on a charge q due to many other charges, Q1, Q2, ...Qn is just the sum of the forces due to each of these charges, ie. ~Ftot = n∑ i=1 k Qiq r2i r̂i (2) This looks simple, but of course it is a vector sum, so the math can get messy. ri is the distance between charge i and the charge q. r̂i is a unit vector along the vector which goes from charge q to charge Qi. The principle of superposition also applies when there is a continuous distribution of charge. For example charge distributions on rods, discs, spheres etc. However when treating these distibutions, the sum in Eq. (5) becomes an integral. In treating these problems, we define a small element of charge dQ. This is the amount of charge in a small part of the continuous charge distribution. We shall consider three cases: - Lines: Then dQ = λdx, where λ is the linear charge density. - Surfaces: Then dQ = σdA, where σ is the surface charge density. - Volumes: Then dQ = ρdV , where ρ is the volume charge density. 1 Example - uniform ring of charge Consider a thin perfectly circular ring centered at the origin, with radius R, lying in the x-y plane and being uniformly charged with linear charge density λ. Find the force on a charge q placed at position, ~r = (0, 0, z). Solution By symmetry, the x and y components of the force on the charge are zero. We are thus left with the task of calculating the z-component. The distance, r, from the charge to any point on the ring is given by, r2 = R2 + z2 (3) If we define θ to be the angle that the vector ~r makes with the z-axis, then, cos(θ) = z/r (4) The total force on the charge is then, Fz = ∫ 2π 0 kqdQ r2 cos(θ) = kq R2 + z2 z r ∫ 2π 0 λRdφ = kqQz (R2 + z2)3/2 (5) where we used Q = 2πRλ C. The electric field Definition: Electric field = force on a unit charge The electric field at a point ~r due to a charge distribution is defined in terms of the force on a positive test charge, q, placed at position ~r. The precise definition is, ~E(~r) = Limq→0 ~Fq(~r) q (6) We can think of the electric field as the force per unit positive charge at position ~r. Notice that once we have found ~E(~r), we can find the force on any charge, q, by using ~F (~r) = q ~E(~r). Since the Electric field is basically a force, superposition applies to the electric field. All of the properties of the electric field can be derived from Eq. () and superposition, as we now show. First, from the definition of the electric field, we write Coulomb’s law as, ~E(~r) = k Q r2 r̂ (7) 2
