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UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lect. Notes 4 Prof. Steven Errede LECTURE NOTES 4 Work & Electrostatic Energy: Consider the following situation: A stationary / fixed configuration of source charges is used to generate a (net) electric field E ( r ) . A test charge QT is moved from point a to point b in this electric field E ( r ) . How much mechanical work W is done on the test charge QT in moving it (slowly) from point a to point b? At any point r along the path a → b, the electrostatic force acting on QT is: FC ( r ) = QT E ( r ) b q1 · q2 · The mechanical force required to counter/balance the electrostatic force is: E (r ) q3 · a · → · · → · · → · · → · · → qN · Fmech ( r ) = − FC ( r ) = −QT E ( r ) QT @ P ẑ r Fmech E (r ) FC Ο (Net E-field) ŷ x̂ The mechanical work done on the test charge QT along the path a → b is: b b a a W = ∫ Fmech ( r )idl = −QT ∫ E ( r )idl b But: E ( r ) = −∇V ( r ) OR: ΔVab ≡ V ( b ) − V ( a ) = − ∫ E ( r )idl a ∴ W = QT ⎡⎣V ( b ) − V ( a ) ⎤⎦ = QT ΔVab The mechanical work done on the test charge QT along the path a → b: Fmech ( r ) = − FC ( r ) = −QT E ( r ) b b b W = QT ΔVab = QT ⎡⎣V ( b ) − V ( a ) ⎤⎦ = ∫ Fmech ( r )idl = − ∫ FC ( r )idl = −QT ∫ E ( r )idl a a a Now, if point a is the reference point ra = ∞ where V ( ra ) = V ( ∞ ) = 0 and point rb = r ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. 1 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lect. Notes 4 Prof. Steven Errede Then: W = QT ⎡V ( r ) − V ( ∞ ) ⎤ = QT V ( r ) ⎣ ⎦ i.e. the test charge QT is (slowly) mechanically moved from the point ra = ∞ to the point rb = r . Then W = QT V ( r ) is the mechanical work done on the charge QT in moving it along the path a → b from ra = ∞ to rb = r . Thus, now we also see that W = work done on charge QT is also equal to the POTENTIAL ENERGY, P.E. = W = QTV ( r ) P.E. is linearly proportional to the potential V(r) (when referenced to V (ra = ∞) ). POTENTIAL ENERGY = amount of work W it takes to create the system (Joules). POTENTIAL DIFFERENCE = work W it takes to create the system per unit charge (Joules/Coulomb = Volts). UNITS OF WORK, W = UNITS OF POTENTIAL ENERGY, P.E. = JOULES (SI UNITS) W = P.E.= QT V ( r ) = Coulomb-Volts = Joules i.e. 1 Coulomb x 1 Volt = 1 Joule Fundamental unit of electric charge: 1e = 1.602×10−19 Coulombs = 1.602×10−19 C ∴ 1 electron volt = 1.602×10−19 Joules ← energy conversion factor for eV ⇔ Joules = eV Example 1: An electron is initially infinitely far away from a proton, and is initially at rest (as is the proton). Assume the proton is fixed / rigidly attached to the head of a pin at the origin Ο( x, y, z ) = (0, 0, 0) . The electron is “released” at zero velocity and is attracted to the proton. What is its kinetic energy and speed when it is a distance of 1 meter from the proton? Use Energy Conservation: INIT FINAL ETOT = ETOT =0 INIT ETOT = KEeINIT + PEeINIT = 0 FINAL ETOT = KEeFINAL + PEeFINAL = 0 2 KEeINIT = 12 me vinit =0 KEeFINAL = 12 me v 2final ≥ 0 vinit = 0 2 v final = ?? ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Prof. Steven Errede PEeFINAL = QeV p ( rb = 1m ) where V p ( r ) = PEeINIT = QeV p ( ra = ∞ ) Qe=−e Lect. Notes 4 = −eV p ( ra = ∞ ) = −eV p ( rb = 1m ) 1 e2 =− =0 4πε o ∞ =− So: INIT FINAL ETOT = 0 = ETOT = KEeFINAL + PEeFINAL = 0 Or: KEeFINAL = − PEeFINAL = + 1 Qp , 4πε o r with Qp=+e. 1 ⎛ e2 ⎞ ⎜ ⎟<0 4πε o ⎝ 1 ⎠ 1 ⎛ e2 ⎞ ⎜ ⎟ 4πε o ⎝ 1 ⎠ r = 1 meter −19 e = 1.602x10 coulomb ε o = 8.85x10−12 Farads/meter 1.6 ×10−19 ) ( 1 = Joules 4π ∗ 8.85 × 10−12 1 2 ∴ KEeFINAL KEeFINAL = 2.3 × 10−28 Joules But: 1 eV = 1.6 ×10−19 Joules 1 eV 1.6 ×10−19 1 Joule = 6.242×1018 eV OR: 1 Joule = ∴ KEeFINAL = 1.437 × 10−9 eV Now KEeFINAL = 1 me ve2 FINAL = 2.3 ×10−28 Joules 2 But me = 9.11×10−31 kg 1 Joule = 1 Newton-meter b W = ∫ Fmech idl a veFINAL = veFINAL 2 KEeFINAL 2 ∗ 2.3 ×10−28 Joules 2 = = 504.94 ( m / s ) −31 me 9.11×10 kg 22.5 m/sec at r = 1m 22.5 m / s speed of light NOTE: veFINAL ( r = 1m ) c (22.5 m/s) (c = 3×108 m/s) ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. 3 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 KEe(r) > 0 1 = me ve2 ( r ) 2 Lect. Notes 4 Prof. Steven Errede ETOT ( r ) = KEe ( r ) + PEe ( r ) = 0 i.e. KEe ( r ) = − PEe ( r ) ETOT=0 r 1 1 ⎛ e2 ⎞ 2 me ve ( r ) = + ⎜ ⎟ 2 4πε o ⎝ r ⎠ PEe(r) < 0 =− 1 ⎛ e2 ⎞ ⎜ ⎟ 4πε o ⎝ r ⎠ 1 ⎛ e2 ⎞ ⎜ ⎟ 2πε o ⎝ me r ⎠ ve ( r ) = Example 2: An electron is located at x = − ∞ and has an initial velocity v = 22.5 m / s xˆ and is initially infinitely far away from another electron, which is rigidly fixed at the origin Ο( x, y, z ) = ( 0, 0, 0 ) at rest. veINIT = 22.5 m / s 1 e1 e2 x = −∞ Ο = ( 0, 0, 0 ) x̂ What is the distance of closest approach of e1 to e2? Why is there a minimum distance? TOT TOT We know that EeINIT = EeFINAL (Energy is conserved in this process) 1 1 TOT EeINIT = KEeINIT + PEeINIT 1 1 1 TOT EeFINAL = KEeFINAL + PEeFINAL 1 1 1 1 1 e2 2 me vINIT + 0 = 2.3 ×10−28 Joules = 0+ 2 4πε o r 2 22.5 m/s repulsive closest approach Solve for rclosest : = approach (1.6 ×10 ) = 1 meter ! 1 ⎛ e2 ⎞ 1 = = ⎜ ⎟ 2 2πε o ⎝ me vinit 2π ∗ 8.85 × 10−12 9.11× 10−31 × ( 22.5 )2 ⎠ −19 2 ∴ rclosest approach 4 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lect. Notes 4 Prof. Steven Errede INIT FINAL ETOT ( r ) = ETOT = ETOT = 2.3 × 10−28 Joules = constant = KE(r)+PE(r) 2.3×10−28 Joules KE(r) ETOT(r) KE(r) or: PE(r) PE(r) 0m 0 Joules r = 1m r Classical turning point What happens: e1 comes in from r = ∞ , is decelerated and stopped at r =1m (= distance of closest approach), and then e1 is accelerated back out to r = ∞ . vefinal = 22.5 m / s at r = ∞ again. 1 How much mechanical work is done in taking a charge QT around a closed path / closed contour C in an external electric field E ( r ) ? ∫ a a = −QT ∫ W= ∫ ( − F ( r ) )idl = − ∫ Fmech ( r )idl = a a C C C a a a a FC ( r )idl C E ( r )idl = 0 because: C ∫ E ( r )idl = 0 NO work is done on closed contour, because the electrostatic Coulomb Force, FC ( r ) is a conservative force. ∴ Mechanical work done on test charge QT is independent of the path: E (r ) E (r ) E (r ) b QT path1 path2 a Note: ∫ C a a QT =∫ b a +∫ E ( r ) is produced by some (arbitrary) source charge distribution. a b path1 path2 path1 + path2 are arbitrary because FC ( r ) is conservative (Coulomb Pot’l is 1/r, central pot’l no θ , ϕ dependence.) FC ( r ) = −QT ∇V ( r ) ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. 5 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lect. Notes 4 Prof. Steven Errede ELECTROSTATIC ENERGY OF ASSEMBLY OF A POINT CHARGE DISTRIBUTION How much work does it take to assemble a collection of point charges – bringing them in from infinity, one by one? Bringing in the first charge q1 takes NO work (W1 =0), since there is no electric field present, initially. Now bring in the 2nd charge q2 from infinity. The work done in bringing in q2 from infinity is: W2 = q2 ⎛ q1 ⎞ ⎜ ⎟ = q2V1 ( r2 ) 4πε o ⎝ r12 ⎠ with V1 ( r2 ) = 1 ⎛ q1 ⎞ ⎜ ⎟ 4πε o ⎝ r12 ⎠ where r12 = r2 − r1 and r12 = r12 = r2 − r1 = separation distance between r2 & r1 (i.e. between charges q1 & q2) step 2: ∞ step 1: ∞ ẑ q2 r12 q1 r2 r1 Ο ŷ x̂ Now bring in the 3rd charge from infinity. This requires work W3 = q3V1,2 ( r3 ) where V1,2 ( r3 ) is the (total) potential due to charges q1 & q2: 1 ⎛ q1 q2 ⎞ V1,2 ( r3 ) = V1 ( r3 ) + V2 ( r3 ) = ⎜ + ⎟ 4πε o ⎝ r13 r23 ⎠ ∴ W3 = q3V1,2 ( r3 ) = q3 ⎛ q1 q2 ⎞ ⎜ + ⎟ 4πε o ⎝ r13 r23 ⎠ Similarly for the 4th charge: V1,2,3 ( r4 ) = V1 ( r4 ) + V2 ( r4 ) + V3 ( r4 ) = ∴ W4 = q4V1,2,3 ( r4 ) = 6 1 ⎛ q1 q2 q2 ⎞ + ⎟ ⎜ + 4πε o ⎝ r14 r24 r34 ⎠ q4 ⎛ q1 q2 q3 ⎞ + ⎟ ⎜ + 4πε o ⎝ r14 r24 r34 ⎠ ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lect. Notes 4 Prof. Steven Errede The total work necessary to assemble the first 4 charges is thus: WTOT = W1 + W2 + W3 + W4 = 0 + q2V1 ( r2 ) + q3V1,2 ( r3 ) + q4V1,2,3 ( r4 ) = 1 ⎛ q1q2 q1q3 q1q4 q2 q3 q2 q4 q3 q4 ⎞ + + + + + ⎜ ⎟ 4πε o ⎝ r12 r13 r14 r23 r24 r34 ⎠ We can generalize this relation for N charges as: WTOT = 1 4πε o N ∑ i =1 ⎛ qi q j ⎜ ∑ ⎜ rij j =1 j ≠i ⎝ N ⎞ 1 ⎟= ⎟ 8πε o ⎠ j >i N ∑ i =1 ⎛ qi q j ⎜ ∑ ⎜ rij j =1 j ≠i ⎝ N so we don’t count same pair twice! WTOT ⎛ 1 N ⎜ N 1 ⎛ qj ⎜ = ∑ qi ∑ 2 i =1 ⎜⎜ j =1 4πε o ⎜⎝ rij ⎝ j ≠i ⎞ ⎟ ⎟ ⎠ double-counts pairs but factor of 8 (vs. 4) takes care of this! ⎞⎞ 1 N ⎟ ⎟ = ∑ qiV ( ri ) ⎟ ⎟⎟ 2 i =1 ⎠⎠ = V ( ri ) Notice WTOT can be < 0 or can be > 0, depending on signs of qi & qj e.g. if all qiqj terms > 0, WTOT > 0 (for pair-wise repulsive forces) if all qiqj terms < 0, WTOT < 0 (for pair-wise attractive forces) ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. 7 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lect. Notes 4 Prof. Steven Errede ELECTROSTATIC ENERGY OF A CONTINUOUS CHARGE DISTRIBUTION For a discrete / discretized charge distribution: WTOT ⎛ 1 N 1 N ⎜ N 1 ⎛ qi = ∑ qiV ( ri ) = ∑ qi ∑ ⎜ 2 i =1 2 i =1 ⎜⎜ j =1 4πε o ⎜⎝ rij ⎝ j ≠i ⎞ ⎞⎟ ⎟⎟ ⎟ ⎠ ⎟⎠ N ∑ q V (r ) ⇒ ∫ For a continuum volume charge density, ρ ( r ) : i =1 WTOT = i i V dqV ( r ) = ∫ ρ ( r ) V ( r ) dτ V 1 ρ ( r ) V ( r ) dτ 2 V∫ For a continuum line charge density, λ ( r ) : WTOT = 1 λ ( r ) V ( r ) dl 2 C∫ () For a continuum surface charge density, σ r : WTOT = 1 σ ( r ) V ( r ) dA 2 ∫S Using Gauss’ Law: ρ ( r ) = ε o ∇i E WTOT = ε 1 ρ ( r ) V ( r ) dτ = o ∫ 2 V 2 ∫ ( ∇i E )V ( r ) dτ V Now use integration by parts to transfer the derivative from E to V (and also use the divergence theorem): ∫ ∇i( f A) dτ = ∫ f ( ∇i A) dτ + ∫ Ai( ∇ f ) dτ = ∫ Or: ∫ f ( ∇i A ) dτ = − ∫ Ai( ∇ f ) dτ + ∫ fAida V V V Thus: WTOT = 8 V V S fAida (see Griffith’s eqn. 1.59, p. 37) S εo ⎡ − ∫ E ( r )i( ∇V ( r ) ) dτ + ∫ V ( r ) E ( r )idA⎤ 2 ⎣ V S ⎦ = −E ( r ) ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lect. Notes 4 Prof. Steven Errede εo ⎡ − ∫ E ( r )i( ∇V ( r ) ) dτ + ∫ V ( r ) E ( r )idA⎤ WTOT = 2 ⎣ V ⎦ S = εo ⎡ = εo ⎡ 2 ∫ E ( r ) dτ + ∫ V ( r ) E ( r )idA⎤ 2 ⎣ ∫V 2 ⎣ ( E ( r )i E ( r ) ) dτ + ∫ V ( r ) E ( r )idA⎤⎦ S V ⎦ S Now integration volume V and enclosing surface of integration S are arbitrary, as long as all charges are contained within the volume V and enclosed by the surface S. ⎛S⎞ So we can take ⎜ ⎟ → all space (i.e. ∞ - volume & surface) without any loss of generality. ⎝V ⎠ Then the surface integral over all space ∫ V ( r ) E ( r )idA vanishes, because for a localized S All Space charge distribution i.e. one which has finite spatial extent {with characteristic size ~ d}, far away from the localized charge distribution, if there is a net electric charge associated with the localized charge distribution, then V ( r d ) ~ 1 r and E ( r d ) = −∇V ( r d ) ~ 1 r 2 . If the localized charge distribution has no net electric charge, then for r d the potential V ( r d ) and hence the electric field E ( r d ) will fall of more rapidly than 1 r and 1 r 2 , respectively. {Please see/read P435 Lecture notes 8 (and/or Griffiths 3.4, p. 146-159) for further details/understanding of these large-r dependencies.} Thus, we see that both V ( r = ∞ ) = 0 and E ( r = ∞ ) = 0 , hence ∴ WTOT = εo 2 ∫ ∫ S All Space E 2 ( r ) dτ V ( r ) E ( r )idA = 0 and thus: (n.b. ≥ 0 always!) V all space However, note that for a single point charge, since E ( r ) = q Then: WTOT = εo q2 2 ( 4πε o ) ∫ 2 V all space 1 2 q2 r dr sin d d θ θ ϕ = r4 8πε o ∫ ∞ 0 q 1 rˆ 4πε o r 2 1 dr = ∞ !!! r2 The electrostatic energy associated with a point electric charge is infinite (i.e. singular)! The origin r = 0 cannot truly be included here – this formula breaks down – because it mathematically describes only classical physics – and e.g. we know that quantum mechanics takes over at distance scales of: r≤ e = c me c 2 197 MeV − fm 0.511 385.5 fm 385.5 ×10−15 m ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. 9 UIUC Physics 435 EM Fields & Sources I e = λe h , c 197 MeV − fm 2π me c 2 = 511 keV = 0.511 MeV = and Lect. Notes 4 Prof. Steven Errede c = “reduced” Compton wave length of electron, here h = Planck’s constant, me c 2 = 2π Fall Semester, 2007 (1 MeV = 10 eV ) , electron rest mass energy, 6 And 1 fm = 1 femto − meter = 1 fermi = 10−15 m Actually, we know that for atoms & molecules, quantum mechanics takes over at distance scales of ~ Bohr radius (H-atom): ao = 0.53Þ= 0.53 ×10−10 m = 0.053 nm 385.5 ×10−15 m e = 385.5 fm c q The classical formula for WTOT breaks down / is invalid for point charges, for r ≤ e = 385.5 fm me c 2 At an even shorter distance scale, that of the so-called Planck distance LP = with corresponding Planck time t P = LP c = GN c 5 GN c 3 1.6 × 10−35 m , 5.4 ×10−44 s , the “classical”, continuous nature of space-time itself becomes “foam-like” – thus, we certainly have no reason to believe in the validity of Coulomb’s law below the Planck distance scale. WTOT = = εo 2 εo 2 ∫ E ( r )i E ( r ) dτ (Joules) This formula works fine/is valid for continuous charge distributions: λ ( r ) ,σ ( r ) , ρ ( r ) V all space ∫ E 2 ( r ) dτ > 0 (always) V all space E ( r ) is total/net electric field arising from charge distribution, ρ ( r ) . Define: uE ( r ) ≡ Joules/meter3 WTOT = ∫ εo ε E ( r )i E ( r ) ) = E ( 2 2 o u E ( r ) dτ = V all space = 1 2 εo 2 ∫ 2 E 2 ( r ) dτ 1 ( r ) = ρ ( r )V ( r ) 2 = electrostatic energy density = Physically, says that the energy density, uE ( r ) = ε2o E 2 ( r ) V all space is associated entirely with the electric field, E ( r ) . ∫ ρ ( r )V ( r ) dτ Physically, says that the energy density, u E ( r ) = 12 ρ ( r ) V ( r ) V all space is due to product (i.e mix) of volume charge density, ρ ( r ) and electrostatic potential, V ( r ) . Electrostatic energy density: uE ( r ) = 10 εo 2 E2 (r ) = 1 ρ ( r ) V ( r ) (Joules/meter3) 2 ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lect. Notes 4 Prof. Steven Errede ELECTROSTATIC ENERGY AND THE SUPERPOSITION PRINCIPLE Note that since W (work) / P.E. (potential energy) is quadratic in the electric field, i.e. W= εo 2 ∫ E 2 ( r ) dτ = εo ∫ E ( r )i E ( r ) dτ 2 V all space V all space then for example, if we double E , i.e. E → 2 E , the work needed / done to assemble the nowdoubled charge distribution / stored potential energy in the electrostatic field quadruples (i.e increases 4x). Thus, work done to assemble the charge distribution / potential energy stored in the electrostatic field of the charge distribution does NOT obey the principle of linear superposition. This is not surprising, as there exist many other physical examples where the superposition principle does not hold – e.g. the superposition of two (or more) acoustic / sound waves, or e.g. the superposition of two (or more) light waves – sound/light overall intensities (proportional to square of overall/total amplitude) result from interference effects at the amplitude level. Here, e.g. let ETOT ( r ) = E1 ( r ) + E2 ( r ) Then: WTOT = P.E. = = = εo 2 ∫ ( E ( r )i E ( r ) ) dτ TOT εo 2 ∫ ( E ( r ) + E ( r ) )i( E ( r ) + E ( r ) ) dτ 1 2 1 2 V all space εo 2 TOT V all space ∫ ( E ( r ) + 2 E ( r )i E ( r ) + E ( r ) ) dτ 2 1 1 2 2 2 V all space WTOT = P.E. = W1 + W2 + ε o ∫ ( E ( r )i E ( r ) ) dτ 1 2 V all space Clearly, the principle of linear superposition is not (always) obeyed for WTOT – there exists an additional cross-term involving the product the two electric fields: E1 ( r )i E2 ( r ) . ©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005 - 2008. All rights reserved. 11