Electric Fields: Coulomb's Law, Point Charges, and Dipoles - Prof. Massimiliano Bonamente, Study notes of Physics

Download Electric Fields: Coulomb's Law, Point Charges, and Dipoles - Prof. Massimiliano Bonamente and more Study notes Physics in PDF only on Docsity! CHAPTER 22 The electric field Coulomb’s Law describe the force exerted by a charge Q on another charge q. The force is an action at a distance, since no actual connection is required between the two charges. This concept can be described by saying that charge Q (or a set of charges) sets up an electric field E in the space around it, so that a charge will experience a force F given by E = F q (1) It is clear that the electric field E depends on (a) the distribution of charges that set up the field and (b) the location of the test-charge q which experiences the force. Units of measure for the electric field is N C−1. The only meaning associated with the electric field E is that a test charge q immersed in that field will experience a force equal to F = qE (2) 1. Electric field due to a point charge In Figure 1 are shown the so-called electric field lines, or imaginary lines which indicate the direction of the force experienced by a test charge located at that position. It is clear that, if the test charge has a negative charge, then the force would be directed in the opposite direction. The electric field is radial (in the direction of the unit radial vector r̂), and the magnitude of the field at a distance of r from the central point charge is E(r) = k Q r2 r̂ = 1 4πǫ Q r2 r̂ (3) If the central charge Q is positive, the electric field is positive, meaning that a positive charge will experience a positive force F = q · E away from the central charge - in fact, like charges repel. – 2 – + F Fig. 1.— Electric field lines by a positive point charge 2. Electric field due to an electric dipole A dipole is a configuration of a positive charge q and negative charge −q separated by a distance d (Figure 2). The field caused by this configuration of charge can be obtained by linear superposition, although such calculation is complicated by the vector nature of the electric force (and therefore of the electric field; see Figure 22-5 of textbook). The electric field near an electric dipole can be calculated easily in two cases: (1) For points P along the dipole axis. Since the action of the two charges are of opposite sign: E = E+ + E− = k · ( q r2+ − q (r+ + d)2 ) (4) For points at a large distance from the dipole, still along the dipole axis and at r+ >> d, one can simpify the above expression using the following mathematical expansion known as the Binomial theorem: (1 + x)n = n ∑ i=0 ( n i ) xn = 1 + nx 1 + n(n − 1)x2 2! + . . . (5) In this case, rewrite Eq. 4 by retaining only terms of the first order in the binomial expansion: E = kq(r−2+ −(r++d)−2) = kq ·r−2+ [ 1 − ( 1 + ( d r+ )) −2 ] ≃ kq ·r−2+ (1−(1− 2dr+ )) = k · 2·dq r3 + – 5 – ⇒ E = k zλ · 2πR (z2 + R2)3/2 = 1 2ǫ · zλ · R (z2 + R2)3/2 = 1 4πǫ qz (z2 + R2)3/2 (10) where the ring’s total charge is q = 2πRλ. At large distance from the ring, one can approximate z2 + R2 ≃ z2 and E =≃ 1 4πǫ q z2 . This result was to be expected, since at large distances from the loop, the loop behaves as a point charge which behaves according to Coulomb’s law. 3.2. A charged disk A charged disk of surface density σ [C m−2] can be thought of as a sequence of loops of charge dq = σ · 2πxdx, where 2πxdx is the loop’s area, and xǫ[0, R]. This makes it possible to use the results from the charged loop above, and find that the electric field due to a disk of radius R is E = ∫ R 0 k z·σ·2πxdx (z2+x2)3/2 = kπ · zσ · ∫ R 0 (z2 + x2)−3/22xdx E = kπσz (z2 + x2)−1/2 −1/2 ∣ ∣ ∣ ∣ R 0 = 2kπσz ( 1 z − z√ z2 + R2 ) = σ 2ǫ ( 1 − z√ z2 + R2 ) (11) 3.3. An infinite sheet of charge An infinite sheet of charge can be tought of as a charged disk (Eq. 11) in which R → ∞ E = σ 2ǫ (12) The direction of the field is away from the sheet, just like in the case of the loop and the disk. 4. Point charges and dipoles in an electric field The reason for the definition of the electric field is Eq. 1, which can be used to determine the force acting on a charge q in an electric field E. – 6 – The action on an electric dipole is more complex. Consider the force on each charge constituting an elementary dipole: θ Ε − + C p Fig. 4.— A dipole in a uniform electric field |F | = |F+| = |F−| = q · E, with opposite sign, which cause a torque with respect to the center of the dipole: τ = 2 · (d/2F · sin(θ)) τ = p × E (13) Equation 13 is the reason for the definition of the dipole moment p. Notice that in Figure 4 the direction of τ is into the page, since the rotation experienced by the two charges in the dipole is clock-wise, or with negative sign. 4.1. Small oscillations of a dipole in a fixed electric field Consider Figure 4, and a dipole located at a small angle θ, with moment of inertia I. It is clear that the torque exerted on the dipole is given by τ = −pE · sin(θ) ≃ −pE · θ where the negative sign indicates a clock-wise rotation. Recall that an extended object with rotational inertia I subject to a torque τ develops an angular acceleration α = τ/I α = −pE I · θ (14) It is clear, then, that the angular frequency of the oscillations is given by ω2 = pE I , and the frequency by f = 1 2π · √ pE I Hz. – 7 – 4.2. Potential energy of a dipole An electric dipole will naturally have its dipole moment p aligned to the external electric field E. Considering Figure 4, the equilibrium configuration will have θ=0. When the dipole is not aligned, an external work W will have to be done on the dipole (from the external electric field) to bring it from the rest position to a given angle θ: ∆UP = −W = ∫ θ 0 τdθ ∆UP is the potential energy associated with the work done by the conservative elec- trostatic force, which causes the torque τ . The infinitesimal work done by the electrostatic force is dW = τdθ = −pE · sin(θ)dθ. ⇒ ∆UP = pE · (− cos(θ)|θ0) = −pE(·cos(θ) − ·cos(0)). This result can be summarized as Up = −p · E (15) where ‘·’ is the dot product operator, and Up is the potential energy possessed by a dipole of dipole moment p in an external electric field E. Example: Find the work required to rotate a dipole of p = 10−6 C m from -180 degrees (counter-aligned) to -90 degrees (perpendicular) relative to an external electric field of E = 103 N/C. W = −∆Up = −pE(cos(180) − cos(90)) = −pE(−1 + 0) = pE = 10−3 J. Notice that the work is positive, i.e., the external electric field. does positive work on the dipole. Accordingly, the change in potential energy of the dipole is negative, ∆Up = −10−3 J: in fact, the dipole loses its ‘potential’ to do work, as it tends towards its rest position. 5. The Millikan oil-drop experiment This is a classic experiment that was devised by Robert Millikan at the University of Chicago in 1908, and published in 1909. The experiment was designed to measure the elementary charge of the electron. Consider Figure 5: Oil drop become negatively charged as they are introduced into the chamber. One at a time, They are allowed to drift to a lower chamber, where an electric field E has been set up. When the oil drops are observed to float, it means that the electrostatic force balances the weight: