Physics 121 - Electricity and Magnetism Lecture 05 -Electric Potential Y&F Chapter 23 Sect. 1-5 • • • • • • • • • Electric Potential Energy versus Electric Potential Calculating the Potential from the Field Potential due to a Point Charge Equipotential Surfaces Calculating the Field from the Potential Potentials on, within, and near Conductors Potential due to a Group of Point Charges Potential due to a Continuous Charge Distribution Summary 1 Copyright R. Janow Spring 2014 Electrostatics: Two spheres, different radii, one with charge Connect wire between spheres, then disconnect it Initially Q10= 10 C cm r = 10 1 Q1f= ?? wire Q2f= ?? •Are final charges equal? •What determines how charge redistributes itself? Q20= 0 C r2= 20 cm Mechanical analogy: Water pressure Open valve, water flows What determines final water levels? gy = PE/unit mass P2 = rgy2 P1 = rgy1 X Copyright R. Janow Spring 2014 ELECTRIC POTENTIAL V(r ) Potential Energy due to an electric field per unit (test) charge • Closely related to Electrostatic Potential Energy……but…… • DPE: ~ work done ( = force x displacement) • DV: ~ work done/unit charge ( = field x displacement) • Potential summarizes effect of charge on a distant point without specifying a test charge there (Like field, unlike PE) • Scalar field Easier to use than E (vector) • Both DPE and DV imply a reference level • Both PE and V are conservative forces/fields, like gravity • Can determine motion of charged particles using: Second Law, F = qE or PE, Work-KE theorem &/or mechanical energy conservation Units, Dimensions: • Potential Energy U: Joules • Potential V: [U]/[q] Joules/C. = VOLTS • Synonyms for V: both [F][d]/[q], and [q][E][d]/[q] = N.m / C. • Units of field are [V]/[d] = Volts / meter – same as N/C. Copyright R. Janow Spring 2014 Reminder: Work Done by a Constant Force 5-1: In the four examples shown in the sketch, a force F an object and does work. In all four cases, the force has same magnitude and the displacement of the object is to right and has the same magnitude. Rank the cases in order of the work done by the force on object, from most positive to the most negative. acts on the the the Ds A. B. C. D. E. I, IV, III, II II, I, IV, III III, II, IV, I I, IV, II, III III, IV, I, II F I III F F II F IV Copyright R. Janow Spring 2014 Work Done by a Constant Force (a reminder) DW F Ds = FDs cos The work DW done by a constant external force on it is the product of: • the magnitude F of the force • the magnitude Δs of the displacement of the point of application of the force • and cos(θ), where θ is the angle between force and displacement vectors: Dr F F Dr Dr II I WII = FDr WI = 0 F F Dr Dr III If the force varies in direction and/or magnitude along the path: DW F ds WIII = FDr IV WIV = FDr cos f i Example of a “Path Integral” Result may depend on path Copyright R. Janow Spring 2014 Definitions: Electrostatic Potential Energy versus Potential Recall: Conservative Fields definition • Work done BY THE FIELD on a test charge moving from i to f does not depend on the path taken. • Work done around any closed path equals zero. dU = dW = Fe ds dV dU / q0 (basic definition) Fe = q0E POTENTIAL ENERGY DIFFERENCE: Charge q0 moves from i to f along ANY path POTENTIAL DIFFERENCE: Potential is potential energy per unit charge f Uf Ui DU DW Fe ds = q0 E ds f i Path Integral i DU DW DV = = E Ds q0 q0 Vf Vi DV DW / q0 = E ds f i (from basic definition) ( Evaluate integrals on ANY path from i to f ) Copyright R. Janow Spring 2014 Some distinctions and details D U = q0 D V • • • • • • The field depends on a charge distribution elsewhere). A test charge q0 moved between i and f gains or loses potential energy DU. DU does not depend on path DV also does not depend on path and also does not depend on |q0| (test charge). Use Work-KE theorem to link potential differences to motion Only differences in electric potential and PE are meaningful: – Relative reference: Choose arbitrary zero reference level for ΔU or ΔV. – Absolute reference: Set Ui = 0 with all charges infinitely far apart – Volt (V) = SI Unit of electric potential – 1 volt = 1 joule per coulomb = 1 J/C – 1 J = 1 VC and 1 J = 1 N m • Electric field units – new name: – 1 N/C = (1 N/C)(1 VC/1 Nm) = 1 V/m • A convenient energy unit: electron volt – 1 eV = work done moving charge e through a 1 volt potential difference = (1.60×10-19 C)(1 J/C) = 1.60×10-19 J Copyright R. Janow Spring 2014 Work and PE : Who/what does positive or negative work? 5-2: In the figure, suppose we exert a force and move the proton from point i to point f in a uniform electric field directed as shown. Which statement of the following is true? A. B. C. D. E. f i E Electric field does positive work on the proton. Electric potential energy of the proton increases. Electric field does negative work on the proton. Electric potential energy of the proton decreases. Our force does positive work on the proton. Electric potential energy of the proton increases. Our force does positive work on the proton. Electric potential energy of the proton decreases. The changes cannot be determined. •Hint: which directions pertain to displacement and force? Copyright R. Janow Spring 2014 EXAMPLE: Find change in potential as test charge +q0 moves from point i to f in a uniform field i E f Dx o DU and DV depend only on the endpoints ANY PATH from i to f gives same results •uniform field To convert potential to/from PE just multiply/divide by q0 DVfi = E ds path Fe = q0E DUfi DWfi = F ds path DVfi DUfi / q0 DUfi = q0DVfi EXAMPLE: CHOOSE A SIMPLE PATH THROUGH POINT “O” DVf ,i = DVo,i DVf ,o DVo,i = 0 Displacement i o is normal to field (path along equipotential) DVf ,i = DVf ,o = - E Dx = E | Dx | • External agent must do positive work on positive test charge to move it from o f - units of E can be volts/meter • E field does negative work What are signs of DU and DV if test charge isCopyright negative? R. Janow Spring 2014 Potential Function for a Point Charge • • • • Charges are infinitely far apart choose Vinfinity = 0 (reference level) DU = work done on a test charge as it moves to final location DU = q0DV Field is conservative choose most convenient path = radial Find potential V(R) a distance R from a point charge q : V(R) V VR = E ds along radial path from r = R to ds = r̂dr R q dr 1 q E(r ) = k 2 r̂ E ds = V(R) = kq 2 = kq = k r rR R r q V(R ) = k R R R (23.14) • Positive for q > 0, Negative for q<0 • Inversely proportional to r1 NOT r2 Similarly, for potential ENERGY: (use same method but integrate force) U(r ) QV (R ) = k q.Q R (23.9) • Shared PE between q and Q • Overall sign depends on both signs Copyright R. Janow Spring 2014 Equi-potential surfaces: Voltage and potential energy are constant; i.e. DV=0, DU=0 • No change in potential energy along an equi-potential • Zero work is done moving charges along an equi-potential • Electric field must be perpendicular to tangent of equipotential DV E Ds = -E Ds cos() = 0 and DU = DW = Fe Ds = 0 for Ds along surface • Equipotentials are perpendicular to the electric field lines DV = 0 Vi > Vf DV = 0 Vfi DV = 0 CONDUCTORS ARE ALWAYS EQUIPOTENTIALS - Charge on conductors moves to make Einside = 0 - Esurf is perpendicular to surface E is the gradient of V so DV = 0 along any path on or Copyright Spring 2014 inR.aJanow conductor Examples of equipotential surfaces Uniform Field Point charge or outside sphere of charge Dipole Field Equipotentials are planes (evenly spaced) Equipotentials are spheres (not evenly spaced) Equipotentials are not simple shapes Copyright R. Janow Spring 2014 The field E(r) is the gradient of the potential dV E ds = - E ds cos() •Component of ds on E produces potential change •Component of ds normal to E produces no change •Field is normal to equipotential surfaces •For path along equipotential, DV = 0 ds • Gradient = spatial rate of change dV V V V E=- = i ĵ k̂ - V ds is to equipotential ds x y z f (x, y, z ) Math note : is a " partial" derivative x EXAMPLE: UNIFORM FIELD E – 1 dimension DV E Ds = E Ds Ds E DU = q0 DV = q0EDs = FDs Copyright R. Janow Spring 2014 Potential difference between oppositely charged conductors (parallel plate capacitor) + - • Equal and opposite surface charges • All charge resides on inner surfaces (opposite charges attract) Dx L = E= 0 L Dx DV Vf Vi = E Dx Vf E=0 Example: Find the potential difference DV across the capacitor, assuming: • = 1 nanoCoulomb/m2 • Dx = 1 cm & points from negative to positive plate • Uniform field E 1 10-9 DV = E Dx = E Dx = 10-2 0 Vi DV = 1.13 volts A positive test charge +q gains potential energy DU = qDV as it moves from - plate to + plate along any path (including external circuit) Copyright R. Janow Spring 2014 Comparison of point charge and mass formulas for vector and scalar fields FIELD FORCE VECTORS M mM force/unit mass g ( r ) = G r̂ F(r ) = G 2 r̂ Gravitation (acceleration) r r2 1 Q force/unit charge 1 qQ Electrostatics F(r ) = E(r ) = r̂ r̂ (n/C) 4 0 r 2 4 0 r 2 SCALARS Gravitation POTENTIAL ENERGY Ug (r ) = G mM r 1 qQ Electrostatics Ue (r ) = 4 0 r POTENTIAL Vg (r ) = G M r 1 Q Ve (r ) = 4 0 r PE/unit mass (not used often) PE/unit charge Fields and forces ~ 1/R2 but Potentials and PEs ~ 1/R1 U F= s V E= s Copyright R. Janow Spring 2014 Visualizing the potential function V(r) for a positive point charge (2 D) For q negative V is negative (funnel) V(r) 1/r r r Copyright R. Janow Spring 2014 Conductors are always equipotentials Example: Two spheres, different radii, one charged to 90,000 V. Connect wire between spheres – charge moves Conductors come to same potential Charge redistributes to make it so V1f = V2 f r1= 10 cm V10= 90,000 V. wire Q1f Q2f = Q10 r2= 20 cm Initially: V10 kQ10 = 9 104 Volts = Q10 = 1.0 C. r1 V20= 0 V. Q20= 0 V. Find the final charges: V1f = k[Q 10 Q1f ] kQ1f = V2 f = r1 r2 Find the final potential(s): kQ1f V1f = = r1 r2 -1 Q1f = Q10 (1 ) = 0.33 C. r1 Q2f = Q10 Q1f = 0.67 C. 9 109 0.33 x10 6 = 30,000 Volts = V2f Copyright R. Janow 0.1 Spring 2014 Potential inside a hollow conducting shell Vc = Vb (shell is an equipotential) = 18,000 Volts on surface R = 10 cm Shell can be any closed surface (sphere or not) d c a R Find potential Va at point “a” insidea shell b Definition: DVab Va Vb = E ds b Apply Gauss’ Law: choose GS just inside shell: qenc = 0 E = 0 everywhere inside DVab = 0 Va = Vsurface = Vb = Vc = Vd = 18,000 Volts Potential is continuous across surface – field is not V(r) E(r) Vinside=Vsurf Voutside= kq / r Eoutside= kq / r2 Einside=0 R r R r Copyright R. Janow Spring 2014 Potential due to a group of point charges • Use superposition for n point charges V(r ) = n 1 V = i 4 0 i =1 n r i =1 qi ri • The sum is an algebraic sum, not a vector sum. r r2 r r1 r r1 r2 • Reminder: For the electric field, by superposition, for n point charges E(r ) = Ei n i =1 1 4 0 n i=1 qi 2 r̂i r ri • E may be zero where V does not equal to zero. • V may be zero where E does not equal to zero. Copyright R. Janow Spring 2014 Use Superposition Examples: potential due to point charges Note: E may be zero where V does not = 0 V may be zero where E does not = 0 TWO EQUAL CHARGES – Point P at the midpoint between them EP = 0 d +q VP = +q P by symmetry kq kq kq =4 d/2 d/2 d obviously not zero F and E are zero at P but work would have to be done to move a test charge to P from infinity. Let q = 1 nC, d = 2 m: 9 10 9 10 9 VP = 4 = 18 Volts 2 DIPOLE – Otherwise positioned as above EP 0 d +q -q P but Let q = 1 nC, d = 2 m: VP = obviously EP = 2 kq kq =0 d/2 d/2 kq d2 /4 = 8 kq d2 9 109 10 9 EP = = 8 = 18 V/m (or N/C) 4 Copyright R. Janow Spring 2014 Another example: square with charges on corners a q -q d a P a d -q q a Find E & V at center point P d= a 2/2 EP = 0 by symmetry kq k k VP = i = qi = [q q q q] d i d i ri VP = 0 Another example: same as above with all charges positive EP = 0 by symmetry, again kqi k 4kq 8kq VP = = qi = = = 510 Volts r d a 2 /2 a 2 i i i Another example: find work done by 12 volt battery in 1 minute as 1 ampere current flows to light lamp i E + - DW work done = - DU = - QDV Q = charge moved from to - by current = i Dt = 1 amp x 60 sec = 60 C. DU = QDV = 60 x DV DV = 12 Volts DU = 720 Joules Copyright R. Janow Spring 2014 DW = DU = 720 Joules (from battery) Electric Field and Electric Potential 5-3: Which of the following figures have V=0 and E=0 at the red point? q -q q q A q q q q q -q C B -q q D q -q E Copyright R. Janow Spring 2014 Method for finding potential function V at a point P due to a continuous charge distribution 1. Assume V = 0 infinitely far away from charge distribution (finite size) 2. Find an expression for dq, the charge in a “small” chunk of the distribution, in terms of l, , or r ldl for a linear distributi on dq = d2 A for a surface distributi on rd3 V for a volume distributi on Typical challenge: express above in terms of chosen coordinates 3. At point P, dV is the differential contribution to the potential due to a pointlike charge dq located in the distribution. Use symmetry. dq dV = scalar, r = distance from dq to P 4 0r 4. Use “superposition”. Add up (integrate) the contributions over the whole charge distribution, varying the displacement r as needed. Scalar VP. 1 dq VP = dVP = (line, surface, or volume integral) 4 0 r dist dist 5. Field E can be gotten from potential by taking the “gradient”: V Rate of potential change E = V dV E ds perpendicular to equipotential s Copyright R. Janow Spring 2014 Example 23.11: Potential along Z-axis of a ring of charge z Q = charge on the ring l = uniform linear charge density = Q/2a r = distance from dq to “P” = [a2 + z2]1/2 ds = arc length = adf P dq = lds = ladf dq dV = k r 2 kal kQ V = dV = d f = r 0 r ring r z f y a dq x FIND ELECTRIC FIELD USING GRADIENT (along z by symmetry) As Before V= All scalars - no need to worry about direction Almost point charge formula kQ [ z 2 a 2 ]1 / 2 • As z 0, V kQ/a • As a 0 or z inf, V point charge V kQ (z 2 ) kQz Ez = k̂ = 3 k̂ = 2 k̂ 2 3/2 z 2r z [z a ] • E 0 as z 0 (for “a” finite) • E point charge formula forR. Janow z >> Spring a Copyright 2014 Example: Potential Due to a Charged Rod • A rod of length L located parallel to the x axis has a uniform linear charge density . Find the electric potential at a point P located on the y axis a distance d from the origin. • Start with r [x 2 d2 ]1 / 2 dq = ldx 1 dq 1 ldx dV = = 4 0 r 4 0 (x 2 d2 )1 / 2 • Integrate over the charge distribution L l dx l = ln x ( x 2 d2 )1 / 2 2 2 1 / 2 4 0 (x d ) 4 0 0 V = dV = = • l ln L (L2 d2 )1 / 2 ln(d) 4 0 L0 Check by differentiating d log(x r ) dx for r = [x 2 d2 ]1 / 2 d 1 d(x r ) 1 dr 1 x 1 rx 1 log(x r ) = = (1 )= (1 ) = ( )= dx x r dx xr dx xr r xr r r • Result L (L2 d2 )1 / 2 l V= ln 4 0 d Copyright R. Janow Spring 2014 Example 23.10: Potential near an infinitely long charged line or charged conducting cylinder E= 2kl r Near line or outside cylinder r > R E is radial. Choose radial integration path if DVfi = Vf - Vi = i f f dr E d r = 2kl i r DVfi = 2kl ln[ri / rf ] Above is negative for rf > ri with l positive E=0 Inside conducting cylinder r < R E is radial. Choose radial integration path if DVinside = i f E dr = 0 Potential inside is constant and equals surface value Copyright R. Janow Spring 2014 Example 23.12: Potential at a symmetry point near a finite line of charge l = Q/2a Uniform linear charge density dq = ldy Charge in length dy dq Potential of point charge dV = k r a a dy VP = dV = kl -a - a (x 2 y2 )1/ 2 Standard integral from tables: (x 2 dy = y2 )1/ 2 ln [ y r ] = ln [ y (x 2 y2 )1/ 2 ] VP = VP a (x 2 a2 )1/ 2 kl ln 2 2 1/ 2 a (x a ) (x 2 a2 )1/ 2 a kQ = ln 2 2 1/ 2 2a (x a ) a Limiting cases: • Point charge formula for x >> 2a • Example 23.10 formula for near field limit x << 2a Copyright R. Janow Spring 2014 Example: Potential on the symmetry axis of a charged disk P • Q = charge on disk whose radius = R. • Uniform surface charge density = Q/4R2 • Disc is a set of rings, each of them da wide in radius • For one of the rings: dq dA = a da df VP,z 1 = 4 0 2 R 0 0 dVP ,z a da d [ a 2 z 2 ]1/ 2 r z R dA = adfda cos() = z/r r 2 = a2 z2 dA=adfda k dq = r f a Double integral x • Integrate twice: first on azimuthal angle f from 0 to 2 which yields a factor of 2 then on ring radius a from 0 to R (note: (1 x)1/ 2 1 21 x 21 41 x2..... for x2 1 ) 2 a da 4 0 0 [ a 2 z 2 ]1/ 2 R VP,z = “Far field” (z>>R): disc looks like point charge a d 2 Use Anti= [ a z 2 ]1/ 2 2 2 1/ 2 derivative: [ a z ] da Vdisk = 2 0 z 2 R2 1/2 z Vdisk 2 0 1 R2 1 Q z = z 2 z 4 0 z “Near field” (z<<R): disc looks like infinite sheet of charge Vdisk R 20 z 1 R E dV = dz 2 0 Copyright R. Janow Spring 2014

© Copyright 2018