lect06 - Physics 227: Lecture 6 Dipoles, Calculating...

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Unformatted text preview: Physics 227: Lecture 6 Dipoles, Calculating Potential Energy or Potential, Equipotential Lines • Lecture 5 review: • • • The electric field vanishes inside a spherical shell of charge. Conductors and surface charge. Conservative forces, fields, potential energy, work, potentials: F = -∇U, U = -∫F.dx ➭ UC = kQq/r. Thursday, September 22, 2011 The Dipole • Put a pair of charges, ±q, a fixed distance d apart in a constant E field. • Why are dipoles important? Although molecules are typically neutral, there are some, like H2O, that have a static dipole moment. • The Picture copied out of Wikipedia Thursday, September 22, 2011 dipole moment of water is ≈6x10-30 C.m. If we take one electron for q, then d = p/q ≈ 40 pm, nearly the size of a hydrogen atom (r ≈ 40 pm). The electrons move away from the H towards the O. The Dipole • Put a pair of charges, ±q, a fixed distance d apart in a constant E field. • The total force from the external field is 0... but it generates a torque that causes the dipole to rotate: τ = 2qE(d/2) sinϕ = qEd sinϕ. • The structure of the dipole comes in as a product qd, which we define as the dipole moment: p = qd. • Thus: τ = pE sinΦ, or in vector form: τ = p x E • You can see that the work done by the electric field is dW = 2 q E (d/2) sinΦ (-dΦ) = -pE sinΦ dΦ = pE d(cosΦ). • Thus: Thursday, September 22, 2011 dU = -dW = -pE d(cosΦ), or U=-pE cosΦ = -p.E The Dipole • Put a pair of charges, ±q, a fixed distance d apart in a constant E field. • The total force from the external field is 0... because the field is constant. If the field is not uniform, then we would expect that usually the force is non-zero. Thursday, September 22, 2011 Calculating Potential / Potential Energy • Generally there are two ways to do this. • The electric field E is specified, and you do a line integral of E.dl. • The charge distribution is specified and you do a volume integral of kρ/r or a sum over charges of kqi/r. • Thursday, September 22, 2011 Potential is a scalar, not a vector quantity, and the sums and integrals are similar to those for ˆ determining the electric field: r/r2 → r Single Point Charge Potential, Potential Energy of Two charges • • V = kQ/r • The standard convenient choice is V = 0 at r = ∞. There is no point in adding a constant to this. U = kQq/r • Again we set the potential energy to 0 at r = ∞. Thursday, September 22, 2011 Potential Energy for Two Charges • The potential energy has the same magnitude, but o pposite sign, depending on whether the charges have the same or opposite signs. Thursday, September 22, 2011 Potential from a ring of charge • On the axis, we can get the potential from a ring of charge by integrating: V(r) = ∫k dq / r = kq/r. • The formula is the same result as for a point charge, because on the axis all points on the loop are equally far away and we use the variable r. • Note rminimum = a; r does not go to 0. • With a point charge at O, we have on the x-axis V(x) = kq/x, the same formula as V(r) except for a change of variables. • Whereas for the ring of charge we have V(x) = kq/(x2+a2)1/2. We have the same formula and different limits on r, or the same variable x and different formulas. Thursday, September 22, 2011 Potential iClicker For the three charge distributions shown, which of the choices shown for the orderings of the potential at the point X is correct? You may assume that the charge density per unit length along the circular arcs is the same in all cases. You may assume the charge on the ring is positive. Thursday, September 22, 2011 A. VA = VB = VC. B. VA > VB > VC. C. VA > VB = VC. D. VA = VB > VC. E. none of the others. Potential iClicker As per the ring of charge derivation, the field for each is kqi/r, since all the charge is r away from the central x, but qB = qC while qA is twice as large. So answer C. Thursday, September 22, 2011 A. VA = VB = VC. B. VA > VB > VC. C. VA > VB = VC. D. VA = VB > VC. E. none of the others. Another Potential iClicker For the three charge distributions shown, which of the choices shown for the orderings of the potential at the point X is correct? But now assume that the total charge in each of the 3 loops is the same - B and C have the same charge density, and A has half the charge density of B and C. You may assume the charge on the ring is positive. Thursday, September 22, 2011 A. VA = VB = VC. B. VA > VB > VC. C. VA > VB = VC. D. VA = VB > VC. E. none of the others. Another Potential iClicker A. VA = VB = VC. Since the total charge is not the same for each ring, the potential is the same in each case as well. B. VA > VB > VC. C. VA > VB = VC. D. VA = VB > VC. E. none of the others. Thursday, September 22, 2011 Potential vs. Field for a Charged Spherical Shell • Outside the spherical shell, we know E(r) = kq/r2, and the potential is V(r) = kq/r. • What happens inside the spherical shell? • The electric field is 0 inside the spherical shell. •If E=0, then ΔV = -∫E.dl = 0, so the potential is constant. Thursday, September 22, 2011 Potential Generated by Two Fixed Charges y • When r1 we have a charge > 0 and one < 0, the potential is V(r) = kq1/r1 - kq2/r2. r2 • The +q1 Thursday, September 22, 2011 -q2 x potential is 0 wherever q1/r1 = q2/r2, or r1 = (q1/q2)r2. • If q1 = q2, this defines the y axis. Otherwise we have a closed curve around the smaller magnitude charge. Potential from an infinite Line of charge • Integrating kdq/r looks harder, vs. integrating (λ/2πε0r)dr which looks easier: ∆V = − • As • The ￿ a b λ r · d￿ ˆr 2π￿0 r ∫dr/r → ln(r), we obtain: ￿ r0 ￿ λ ln ∆V = 2π￿0 r potential decreases as r increases. It goes to 0 at some reference radius r0, and then becomes negative. We cannot choose r0 = ∞, as then V = ln(∞) = ∞ everywhere. Thursday, September 22, 2011 Potential from an infinite cylinder of charge • For an infinite cylinder of charge, the potential is the same outside the cylinder. Using λ = πr2ρ, we have again: • ￿ r0 ￿ λ ln ∆V = 2π￿0 r It is convenient to choose r0 = R, the radius of the cylinder. • What happens inside the cylinder? • E = (ρ/2ε0)r, so V= -(ρ/4ε0)r2 + arbitrary constant. To match V=0 at r=R, the constant is (ρ/4ε0)R2. • Thursday, September 22, 2011 Potential for a Parallel PLate Capacitor / infinite Planes of charge + + + + + + + + + + - - - Charge density: ±σ C/m2 on each conducting plane E + + • We assume the plate area A is square and large compared to the plate separation d (√A = l >> d). That is, we ignore fringe fields and treat the plates as infinite in extent, and the E field as constant. • Drawing a Gaussian surface, we see EAG = σAG/ε0 → E = σ/ε0. • The voltage between the plates is V = ∫E.dl = Ed = σd/ε0 = Qd/Aε0. • We will come back to the energy stored in a capacitor next week. Thursday, September 22, 2011 Algebraic Problem Example V = 4x + 3y − 2z 2 3 This potential pulled out of “thin air”. ￿d d d￿ 2 3 ￿ = −∇V = − E x+ ˆ y + z (4x + 3y − 2z ) ˆ ˆ dx dy dz ￿ = −8xx − 3ˆ + 6z 2 z E ˆ y ˆ V = 5xy ￿ = −5y x − 5xy E ˆ ˆ This potential pulled out of “thin air”. Thursday, September 22, 2011 Graphical Problem Example V(x) is shown. Which plot shows the correct E(x)? A. B. C. D. None of them. Thursday, September 22, 2011 Spherical Conductor with Cavity There is a charge +q at the center of the cavity. +q What does the electrical potential look like? V V V C. r Thursday, September 22, 2011 V A. r D. B. r V E. r r F. A-E are all nonsense. E ≈ 1/r2 Spherical Conductor with Cavity 0 There is a charge +q at the center of the cavity. 1/r2 +q What does the electrical potential look like? V V A. B. V ≈ 1/r constant 1/r V V C. r Thursday, September 22, 2011 r D. r V E. r r F. A-E are all nonsense. Equipotential Lines • • Topographic maps shows lines of constant elevation = lines of constant gravitational potential - let’s not even discuss any potential issues due to the earth’s non-sphericity and rotation When the lines are closer together, the slope is greater. Equipotential plots in electrostatics play the same role as the lines of constant elevation in topographic maps. Thursday, September 22, 2011 Equipotential Lines • • • • • The familiar field lines are shown in red. The equipotential lines are shown in blue. Field lines have a direction, whereas equipotential lines do not. Field lines do not touch or cross, but equipotential lines can, where E=0. Each is perpendicular to the other. Why? E = -∇V and the direction of the greatest change in V is the perpendicular to the V=constant line. Thursday, September 22, 2011 Equipotential Lines • A puzzle: How can equipotential lines cross if... • • The electric field is a unique vector at each point in space, and E = -∇V so the field line is the perpendicular to the equipotential line? Thursday, September 22, 2011 Thank you. On Monday, Sep 26 I will be traveling. Prof. Cizewski will be giving the lecture. On Thursday Sep 29 I will still be away. Prof. Chandra will give the lecture. See you Monday Oct 3. Thursday, September 22, 2011 ...
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This note was uploaded on 09/27/2011 for the course PHYSICS 750:227 taught by Professor Ronaldgilman during the Fall '11 term at Rutgers.

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lect06 - Physics 227: Lecture 6 Dipoles, Calculating...

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