1 1. Electric force is conservative force ) 11 ( 44 d d 4 d 4 d ? ddd 0 2 0 2 0 2 0 fi L r r rr qQ r rqQ WW r rqQ r rrqQ rEqrFW f i ?=== = ? =?=?= ∫∫ πεπε πεπε r r r r r Source charge： Test charge：q Q EqF rr = ' rQ r r a r b a r r r r b r r d Eq r L rd + + §17.1 Electrical potential energy and electric potential Coulomb’s force: 21 2 210 ? 4 1 r r qQ F πε = r 2 )() 11 ( 4 ) 11 ( 4 d 0 0 f i if if fi PEPE rr qQ rr qQ rEqW ??=??= ?=?= ∫ πε πε r r ∴ PEPEPEWW if ??=??== )( conserv KEPEWWW ?? =?+=+ )( nonconconservnoncon Q 2. Electrical potential energy The electrical potential energy of charge q r qQ PE 1 4 0 πε = §17.1 Electrical potential energy and electric potential Divided the equation by the charge q )() 11 ( 4 d 0 f i q PE q PE rr Q rE i f if ??=??=? ∫ πε r r 3. The electric potential and electric potential difference The electric potential V at a point in an electric field is the potential energy per unit charge at that point. The electric potential q PE V qof = Define: qVPE = qof the potential energy of q §17.1 Electrical potential energy and electric potential 3 Define the electric potential difference )()(d f i q PE q PE VVrE i f if ??=??=? ∫ r r ∫ ?=? f i drEVV fi r r or Note: 1The electric potential is not same thing as the electric potential energy.Electric potential is a property of a point in space, whether or not a charge is placed at that point. It can be positive, negative or zero. §17.1 Electrical potential energy and electric potential 2In PE=qV, PE depends on both the sign of q and the sign of V at the point where q is placed; 3the electric potential and the electric potential energy are scalar quantities. 4The place where the electric potential and the electric potential energy is set to zero is arbitrary, we can choose it anywhere we like for a finite charge distribution. The unit of the electric potential: volt(v)=J/C §17.1 Electrical potential energy and electric potential 4 ∫ ?=? f i drEVV fi r r r r d ※The first method of calculating the electric potential §17.1 Electrical potential energy and electric potential §17.2 The calculation of the electric potential 1. The electric potential of a pointlike charge Q P r i r q E r r r d fi fi r r VV rr Q rr r Q rE f i ?=?=?=? ∫∫ ) 11 ( 4 d ? 4 d 0 2 0 f i πεπε rr r 5 Choose V f =V(r f )=V(∞)=0, the electric potential of a point charge Q is r Q rV r Q rV i i 00 4 )( 4 )( πεπε == or Q+ x y )(rV Q? §17.2 The calculation of the electric potential (i)It is obviously that when r→∞, V approaches 0. (ii)the electric potential varies inversely the distance r to the first power, and the electric field varies inversely the distance r to the second power. (iii) the electric potential is scalar, the electric field is vector. Note: §17.2 The calculation of the electric potential 6 2. The electric potential of a collection of pointlike charges The total electric potential is the algebraic scalar sum of the potential s of each charges. L L +++= +++= 30 3 20 2 10 1 321 444 r Q r Q r Q VVVV πεπεπε 3. The electric potential of continuous charge distributions of finite size ∫ == distr. charge 00 d 4 1 4 d d r Q V r Q V πεπε §17.2 The calculation of the electric potential Note: This method can be applied only to the finite charge distribution. If the charge distribution is infinite, we must use the integral of the electric field as in the definition of the electric potential. Example 1: P773 17.2 Example 2: P773 17.3 ※The second method of calculating the electric potential ∫ == distr. charge 00 d 4 1 4 d d r Q V r Q V πεπε §17.2 The calculation of the electric potential 7 1Find the electric potential a distance r from a spherical shell of radius R that has a charge Q distributed uniformly throughout its surface, for r >R and r <R respectively. rr Q r rrQ rEV Pr 1 4 4 d ? d 0 2 0 outout ∝= ? =?= ∫∫ ∞∞ πε πε r r r )( 4 ? )( 0 2 0 Rr r rQ Rr E > < = πε r R Q o P r r E r o 2 1 r ∝ r E R 0= ∞ Vchoose §17.2 The calculation of the electric potential constant 44 d ? dd d 0 2 0 outin inside ' ' = = ? = ?+?= ?= ∫ ∫∫ ∫ ∞ ∞ ∞ R Q r rrQ rErE rEV R R R P P πεπε r r r r r r r rr Q V 1 4 0 outside ∝= πε R Q o P r r E r o 2 1 r ∝ r E R P′ r 1 ∝ rRo R Q 0 4πε V §17.2 The calculation of the electric potential 8 2Find the electric potential a distance r from the center of a sphere of radius R that has a charge Q distributed uniformly throughout its volume, for r >R and r <R, respectively. )( 4 ? )( 4 2 0 3 0 ? ? ? ? ? ? ? > < = Rr r rQ Rrr R Q E πε πε r r rr Q r rrQ rEV Pr 1 4 4 d ? d 0 2 0 outoutside ∝= ? =?= ∫∫ ∞∞ πε πε r r r 0= ∞ V choose Q R r O P §17.2 The calculation of the electric potential Q R r O P )3( 24 1 4 ) 22 ( 4 4 d ? 4 d dd d 2 2 0 0 22 3 0 2 0 3 0 outin inside ' ' R r R Q R QrR R Q r rrQ R rrQ rErE rEV R R r R R P P ?= +?= ? + ? = ?+?= ?= ∫∫ ∫∫ ∫ ∞ ∞ ∞ πε πεπε πεπε rrr r r r r r r As shown in Fig.17.20 §17.2 The calculation of the electric potential If we choose V O =0,V inside =? 9 ? ? ? ? ? >< <<? = ) , 0( 0 )0( 0 dxx dx E ε σ §17.2 The calculation of the electric potential 3Calculate the electric potential at a point x between two infinite, uniformly charged plates, separated by a distance d. ? σ+ O x σ? 0region <x 0d0d 00 ==?= ∫∫ xx xrEV r r 0= O V dx >region dxixiExrEV dd d x 0 0 0 0f i d ? d) ? (d0d ε σ ε σ =?=??+=?= ∫∫∫∫ r r dx <<0region xx ixiErEV x x 0 0 0 0f i d ? d) ? (d ε σ ε σ =?= ??=?= ∫ ∫∫ r r ? σ? O x σ+ x V O d EddVdV ==? 0 )0()( ε σ §17.2 The calculation of the electric potential 10 4the electric potential due to an electric dipole )()( )()( 0 )()(0 )()( 2 1 4 )( 4 1 ?+ +? ?+ ?+ = ? = ? += +== ∑ rr rr q r q r q VVVV i i πε πε §17.2 The calculation of the electric potential dr >>If 2 0 2 0 2 )()( )()( cos 4 1 cos 4 cos r p r dq V rrr drr θ πε θ πε θ = = ≈ ≈? +? +? §17.2 The calculation of the electric potential 11 5Find the electric potential at a point P located a distance z along the axis of a uniformly charged circular ring of radius R with total charge Q. ∫ == r Q V r Q V d 4 1 4 d d 00 πεπε Break up the extended charge distribution into pointlike charge elements Qd 0= ∞ Vchoose R O P z z Qd 2122 )( zRr += §17.2 The calculation of the electric potential 2122 0 2122 00 )(4 1 )( d 4 1d 4 1 zR Q zR Q r Q V + = + == ∫∫ πε πεπε (i)Q >0, V >0; Q <0 V <0. (ii)z >>R, V → (iii)z→∞, V →0. (iv)z =0, z Q V 0 4 1 πε = z Q V 0 4 1 πε = §17.2 The calculation of the electric potential 12 6Find the electric potential at a point P located on the axis of a uniformly charged circular disk of radius R and charge Q a distance z from the plane of the disk. Choose the charge element as a differential ring, same center with the disk. rrQ d2d πσ= 2 R Q π σ = r rd r ′ §17.2 The calculation of the electric potential 2122 0 )( d 4 1 d zr Q V + = πε ])[( 2 4 1 ])[(2 4 1 )( d2 4 1 2122 2 0 2122 0 0 2122 0 zzR R Q zzR zr rr V R ?+= ?+= + = ∫ πε πσ πε πσ πε §17.2 The calculation of the electric potential r rd r ′ 13 Exercise 1: 12 electrons are fixed as shown in fig. (a) and (b), what is the electric potential at point C due to these electrons? §17.2 The calculation of the electric potential Solution: (a) R e R e V 00 3 4 12 πεπε ?=?= 0=E r (b) 0≠E r R e R e V 00 3 4 12 πεπε ?=?= Exercise 2: L, λ, d is given, find the electric potential at point P 1 and point P 2 . At point P 1 : x xd λ )ln( 40 )ln( 4 )( d 4 1 d 4 1 d 4 1d 4 1 d 00 0 0 0 0 00 d Ld L xd xd x r x V r x r q V L L + =+= + = = == ∫ ∫ πε λ πε λ λ πε λ πε λ πεπε §17.2 The calculation of the electric potential Solution: 14 At point P 2 : y LyL L xyx xy x r x V r x r q V L L 2122 0 2122 0 0 2122 0 0 0 00 )( ln 4 0 ])(ln[ 4 )( d 4 1 d 4 1 d 4 1d 4 1 d ++ = ++= + = = == ∫ ∫ πε λ πε λ λ πε λ πε λ πεπε x xd λ §17.2 The calculation of the electric potential θθπσ θπσσ cosdtg2 cosd2dd xx xrSq ?= ?== () x xr q V d 2 tg 4 d d 0 2 1 22 0 ε θσ επ = + = () 12 0 tg tg0 2 d 2 tg d 2 1 RRxVV R R ?=== ∫∫ ε σ ε θσ θ θ Exercise 3: Find the electric potential at point O. The charge surface density of a frustum of a cone with radius R 1 and R 2 respectively, is σ. §17.2 The calculation of the electric potential Solution: θ 1 R 2 R σ O x θcos dx r 15 Exercise 4: A plastic disk is charged on one side with a uniform surface charge density σ, and then three quadrants of the disk are removed. What is the potential due to the remaining quadrant at point P? §17.2 The calculation of the electric potential Solution: ])[( 2 )( d2 4 1 2122 0 0 2122 0 zzR zr rr V R ?+= + = ∫ ε σ πσ πε ])[( 84 2122 0 .qua zzR V V ?+== ε σ §17.3 Equipotential volumes and surfaces 1. Equipotential surfaces Adjacent points that have the same electric potential form an equipotential surface, which can be either an imaginary surface or real, physical surface. A point charge r Q V 0 4πε = The equipotential surfaces are spherical surfaces. 16 xx ixiErEV x x 0 0 0 0f i d ? d) ? (d ε σ ε σ =?= ??=?= ∫ ∫∫ r r ? σ? O x σ+ x V O d §17.3 Equipotential volumes and surfaces )()( )()( 0 )()(0 )()( 2 1 4 )( 4 1 ?+ +? ?+ ?+ = ? = ? += +== ∑ rr rr q r q r q VVVV i i πε πε An electric dipole §17.3 Equipotential volumes and surfaces 17 2. The properties of the equipotential surfaces 2the electric field is always point to the falling down direction of the electric potential. 1The electric field is always perpendicular to the equipotential surface . ⊥∴ ?= ?=? ∫ ∫ E rE rEVV C B C B CB r r r r r Q d0 d Equipotential surface r r d §17.3 Equipotential volumes and surfaces rE rEVV A B AB r r r r Q d// 0d ∴ >?=? ∫ r r d 3the work done by the electric field is zero when a charge is moved on the equipotential surfaces. 0d)( =?=? ∫ C B CB rEqVVq r r §17.3 Equipotential volumes and surfaces 18 3. The relationship between the electric field and the electric potential s V E sEsEV s s d d ddd ?= =?=? r r The single s-component of the electric field is the negative of the derivative of the electric potential with respect to the single s-coordinate. §17.3 Equipotential volumes and surfaces Generalize to three dimension: V k z V j y V i x V E k z j y i x z V E y V E x V E zyx ??= ? ? + ? ? + ? ? ?= ? ? + ? ? + ? ? ≡? ? ? ?= ? ? ?= ? ? ?= ) ??? ( ??? operatordel r §17.3 Equipotential volumes and surfaces 19 4. The electric potential and the electric field of a dipole 2 0 )()( )()( 0 cos 4 1 4 r p rr rr q V θ πε πε = ? = ?+ +? 21222 2222 )( cos zyx z zyxr ++ = ++= θ §17.3 Equipotential volumes and surfaces 23222 0 )(4 zyx zp V ++ = πε 25222 2 0 25222 0 25222 0 )( 3 4 )( 3 4 )( 3 4 zyx zp z V E zyx yzp y V E zyx xzp x V E z y x ++ = ? ? ?= ++ = ? ? ?= ++ = ? ? ?= πε πε πε §17.3 Equipotential volumes and surfaces 20 Del operator in polar coordinates ) ? sin ? cos2 ( 4 cos 4 1 ) ? 1 ? ( 33 0 2 0 θ θθ πε θ πε θ θ r r r p r p r r r E += ? ? + ? ? ?= r θ θ ? 1 ? ? ? + ? ? =? r r r Along the axis of the dipole θ =0o r r p E ? 4 2 3 0 πε = r In the equatorial plane of the dipole θ =90o θ πε ? 4 3 0 r p E = r y z r ? r ? θ ? θ ? θ θ ′ §17.3 Equipotential volumes and surfaces §17.4 The interaction of charged particle with the electric field 1. A single charge q in an electric field PEKEW ?? += noncon CWE theorem In a electric field 0 noncon =+= PEKEW ?? VqqVqVPEKE if ??? ?=??=?= )( q >0, ?V <0, ?KE >0; q <0, ?V <0, ?KE <0; q >0, ?V >0, ?KE <0; q <0, ?V >0, ?KE >0. + ? EqF rr = F r F r 21 2. An electric dipole in an external electric field 1The torque on the dipole EQ d F d r r r r r ×=×= 22 11 τ )( 22 22 EQ d F d r r r r r ?×?=×?=τ EpEdQ EQ d EQ d r r rr r r r r rrr ×=×= ×+×= += 22 21 τττ Q+ Q? §17.4 The interaction of charged particle with the electric field 2the energy of the dipole in an electric field EpPE r r ??= dipole Parallel PE dipole =-pE--stable Antiparallel PE dipole =pE--unstable Ep QEd QEx PEPEPE QVPEQVPE ExVV QQ QQ r r ??= ?= ?= += =?= =? +? +? θcos , dipole 21 21 1 V 2 V Q+ Q? x §17.4 The interaction of charged particle with the electric field 22 §17.5 The electric potential energy of a distribution of pointlike charges 1. Bring a point charge Q 1 to an isolated location from infinite location, no work is done, the change of the potential energy 2.bring the second point charge Q 2 near to the charge Q 1 , the change of the potential energy 1 Q 2 Q 12 r 0 4 1 )( 12 21 0 2 2 ?= ?= ?= r QQ VVQ PEPEPE if if πε ? 0 111 =?=?= ifif VQVQPEPEPE? 3. Bring the third charge Q 3 near the both charges Q 1 and Q 2 , the change of the potential energy 1 Q 3 Q 2 Q 12 r 13 r 23 r 0) 4 1 4 1 ( )( 23 2 013 1 0 3 33 ?+= ?=?= r Q r Q Q VVQPEPEPE ifif πεπε ? §17.5 The electric potential energy of a distribution of pointlike charges 23 The total change of the potential energy )( 4 1 23 32 13 31 12 21 0 321total r QQ r QQ r QQ PEPEPEPEPE ++= ++== πε ???? The work done by the electric field )( 4 1 23 32 13 31 12 21 0 totalelec r QQ r QQ r QQ PEPEW ++?= ?=?= πε ? .0,0;0,0 electotalelectotal <>>< WPEWPE §17.5 The electric potential energy of a distribution of pointlike charges Exercise 1: what is the electric potential energy for the collection of the point charges shown in figure? Where Solution: )( 4 1 4 1 0 3231 0 33 21 0 22 11 d QQ d QQ PEPE d QQ PEPE PEPE +== == == πε ? πε ? ? d q d qq d qq d qq PE 0 2 0 4 10 ) )2)(4()2()4( ( 4 1 πεπε ? = ? ++ ? = qQqQqQ 2,4, 321 +=?=+= 1 Q 2 Q 3 Q §17.5 The electric potential energy of a distribution of pointlike charges 24 §17.5 The electric potential energy of a distribution of pointlike charges Exercise 2: Derive an expression for the work required to set up the four charge configuration shown in Figure, assuming the charges are initially infinitely far apart. Solution: choose 0= ∞ V a q a q a aa a aa q PE f 0 2 0 2 0 2 21.0 )2 2 1 ( 4 2 ] 2 111 2 111 [ 4 επε πε ?=?= +??+??= a q PEPEPEW if 0 2 21.0 ε ? ?=?== 0= i U Exercise 3: How much work is required to bring the charge of +5q in from infinity along the dashed line and place it as shown in Figure? §17.5 The electric potential energy of a distribution of pointlike charges 25 §17.5 The electric potential energy of a distribution of pointlike charges Exercise 4: (a) Show that the electric potential of an infinite sheet of charge can be written V=V 0 -(σ/ε 0 )z, where V 0 is the electric potential at the surface of the sheet and z is the perpendicular distance from the sheet. (b) How much work is required be done by the electric field of the sheet as small positive charge is moved from an initial position on the sheet to a final position located a distance z from the sheet? Solution: (a)The electric potential difference between the sheet and the point away from the sheet 0 0 0 0 0 2 d 2 d ε σ ε σ z zrEVV zz ?==?=? ∫∫ r r 0 0 2ε σz VV ?= therefore (b)the work done by electric field of the sheet 0 0 00 2 )( ε σzq VVqW =?= §17.5 The electric potential energy of a distribution of pointlike charges