Copyright R. Janow Fall Physics 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

Copyright R. Janow Fall 2015 Electrostatics: Two spheres, different radii, one with charge wire Connect wire between spheres, then disconnect it Q 1f = ?? Q 2f = ?? Are final charges equal? What determines how charge redistributes itself? r 1 = 10 cm r 2 = 20 cm Q 10 = 10 C Q 20 = 0 C Initially Mechanical analogy: Water pressure Open valve, water flows What determines final water levels? X P 1 = r gy 1 P 2 = r gy 2 gy = PE/unit mass

Copyright R. Janow Fall 2015 ELECTRIC POTENTIAL Units, Dimensions: Potential Energy U: Joules Potential V: [U]/[q] Joules/C. = VOLTS Units of E field are [V]/[d] = Volts / meter – same as N/C. Related to Electrostatic Potential Energy……but……  PE = Work done ( = force x displacement = charge x field x displacement )  V = Work done/unit charge ( = field x displacement ). Change in PE of a test charge as it is displaced / test charge magnitude Potential: Is a scalar field  easier to use than E (vector). E = Gradient (Potential) Summarizes effect of distant charges without specifying a test charge there. Both  PE and  V: Use a reference level Represent 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 Define: Electrostatic Potential … equals … Potential Energy per unit (test) charge

Copyright R. Janow Fall 2015 Reminder: Work Done by a Constant Force 5-1: In the four examples shown in the sketch, a force F acts on an object and does work. In all four cases, the force has the same magnitude and the displacement of the object is to the right and has the same magnitude. Rank the cases in order of the work done by the force on the object, from most positive to the most negative. I II III IV A.I, IV, III, II B.II, I, IV, III C.III, II, IV, I D.I, IV, II, III E.III, IV, I, II DsDs

Copyright R. Janow Fall 2015 Work Done by a Constant Force (a reminder) The work W done by a constant external force on an object 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: II I IIIIV DrDr In general, force may vary in direction and magnitude along the path: Result of a “Path Integration” can depend on path taken

Copyright R. Janow Fall 2015 Definitions: Electrostatic Potential Energy versus Potential POTENTIAL DIFFERENCE: Potential is potential energy per unit charge ( Evaluate integrals on ANY path from i to f ) (from basic definition) E field is “conservative”, like gravity 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. Work done changes potential energy or potential: POTENTIAL ENERGY DIFFERENCE : Charge q 0 moves from i to f along ANY path Path Integral

Copyright R. Janow Fall 2015 Only differences in electric potential and PE are meaningful: –Relative reference: Choose arbitrary zero reference level for ΔU or ΔV. –Absolute reference: Set U i = 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 “Electron volt” is an energy unit: – 1 eV = work done moving charge e through a 1 volt potential difference = (1.60× C)(1 J/C) = 1.60× J Summary distinctions and details The field is created by a charge distribution elsewhere). A test charge q 0 moved between i and f gains or loses potential energy DU.  U and  V do not depend on path.  U also does not depend on |q 0 | (test charge). Use Work-KE theorem to link potential differences to motion.

Copyright R. Janow Fall : 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? Work and PE : Who/what does positive or negative work? A. Electric field does positive work on the proton. Electric potential energy of the proton increases. B. Electric field does negative work on the proton. Electric potential energy of the proton decreases. C. Our external force does positive work on the proton. Electric potential energy of the proton increases. D. Our external force does positive work on the proton. Electric potential energy of the proton decreases. E. The changes cannot be determined. E if Hint: which directions pertain to displacement and force?

Copyright R. Janow Fall 2015 EXAMPLE: Find change in potential as test charge +q 0 moves from point i to f in a uniform field DV or DU depend only on the endpoints ANY PATH from i to f gives same results E i f o DxDx EXAMPLE: CHOOSE A SIMPLE PATH THROUGH POINT “O” Displacement i  o is normal to field (path along equipotential) External agent must do positive work on positive test charge to move it from o  f E field does negative work units of E can be volts/meter What are signs of D U and  V if test charge is negative? To convert potential to/from PE just multiply/divide by q 0 uniform field

Copyright R. Janow Fall 2015 Potential Function for a Point Charge Reference level: infinitely far away choose V(R = infinity) = 0  U = work done on a test charge as it moves to final location  U = q 0  V Field is conservative  choose most convenient path (i.e., radial) Similarly, for potential ENERGY: (use same method but integrate force) Shared PE between q and Q Overall sign depends on both signs Find potential V(R) a distance R from a point charge q : Positive for q > 0, Negative for q<0 Inversely proportional to r 1 NOT r 2

Copyright R. Janow Fall 2015 Visualizing the potential function V(r) for a positive point charge (2 D) For q negative V is negative (funnel) r V(r) 1/r r

Copyright R. Janow Fall 2015 On Equipotential Surfaces: Voltage and potential energy are constant; i.e. V=0, U=0 Zero work is done moving charges along an equipotential and CONDUCTORS ARE ALWAYS EQUIPOTENTIALS - Charge on conductors moves to make E inside = 0 - E surf is perpendicular to surface so DV = 0 along any path on or in a conductor Equipotentials are perpendicular to the electric field lines q V fi V i > V f DV = 0 E is the gradient of V  V = 0 if E is normal to  s

Copyright R. Janow Fall 2015 q The field E(r) is the gradient of the potential 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, D V = 0 EXAMPLE: UNIFORM E FIELD – 1 DIMENSIONAL PATH E DsDs Gradient = spatial rate of change

Copyright R. Janow Fall 2015 Examples of Equipotential Surfaces Uniform Field Equipotentials are planes (evenly spaced) Point charge or outside sphere of charge Equipotentials are spheres (not evenly spaced) Dipole Field Equipotentials are not simple shapes

Copyright R. Janow Fall 2015 Potential difference between oppositely charged conductors (parallel plate capacitor) Assume  x << L (infinite sheets) Equal and opposite surface charges All charge resides on inner surfaces (opposite charges attract) A positive test charge +q gains potential energy  U = q  V as it moves from - plate to + plate along any path (including external circuit) Example: Find the potential difference  V across the capacitor, assuming:  = 1 nanoCoulomb/m 2  x = 1 cm & points from negative to positive plate Uniform field E ++ -- L xx E = 0 VfVf ViVi

Copyright R. Janow Fall 2015 Example: Two spheres, different radii, one charged to 90,000 V. r 1 = 10 cm r 2 = 20 cm V 10 = 90,000 V. V 20 = 0 V. Q 20 = 0 V. wire Connect wire between spheres – charge moves Initially: Find the final charges: Find the final potential(s): Conductors are always equipotentials Conductors come to same potential Charge redistributes to make it so

Copyright R. Janow Fall 2015 Potential inside a hollow conducting shell V c = V b (shell is an equipotential) = 18,000 Volts on surface R = 10 cm Shell can be any closed surface (sphere or not) Find potential V a at point “a” inside shell Definition: Apply Gauss’ Law with GS just inside shell: R a c b d E(r) r E inside =0 R E outside = kq / r 2 V(r) r V inside =V surf R V outside = kq / r Potential is continuous across surface – field is not

Copyright R. Janow Fall 2015 Potential due to a group of point charges Use superposition for n point charges Potential is a scaler… so … The sum is an algebraic sum, not a vector sum. Comparison: For the electric field, by superposition, for n point charges 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 Fall 2015 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 Use Superposition DIPOLE – Otherwise positioned as above +q -q P d TWO EQUAL CHARGES – Point P at the midpoint between them +q P d F and E are zero at P but work would have to be done to move a test charge to P from infinity.

Copyright R. Janow Fall 2015 Another example: square with charges on corners q q -q a d d a a a Find E & V at center point P P Another example: same as above with all charges positive Examples: potential due to point charges - continued

Copyright R. Janow Fall : Which of the following figures have both V=0 and E=0 at the red point? Electric Field and Electric Potential A q q q -q B E q D q q q q q q C

Copyright R. Janow Fall 2015 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) 3. At point P, dV is the differential contribution to the potential due to a point- like charge dq located in the distribution. Use symmetry. 2. Find an expression for dq, the charge in a “small” chunk of the distribution, in terms of, , or   Typical challenge: express above in terms of chosen coordinates 4. Use “superposition”. Add up (integrate) the contributions over the whole charge distribution, varying the displacement r as needed. Scalar V P. 5. Field E can be gotten from potential by taking the “gradient”: Rate of potential change perpendicular to equipotential

Copyright R. Janow Fall 2015 Example 23.11: Potential along Z-axis of a ring of charge x y z r P z a f dq Q = charge on the ring l = uniform linear charge density = Q/2pa r = [a 2 + z 2 ] 1/2 = distance from dq to “P” ds = arc length = ad  FIND ELECTRIC FIELD USING GRADIENT (along z by symmetry) E  0 as z  0 (for “a” finite) E  point charge formula for z >> a As Before All scalars - no need to worry about direction As z  0, V  kQ/a As a  0 or z  inf, V  point charge Looks like a point charge formula, but r is on the ring

Copyright R. Janow Fall 2015 A rod of length L located parallel to the x axis has a uniform linear charge density λ. Find the electric potential at a point P located on the y axis a distance d from the origin. Example: Potential Due to a Charged Rod Above uses the indefinite integral Integrate over the charge distribution on interval 0 < x < L Result Start with

Copyright R. Janow Fall 2015 Example 23.10: Potential near an infinitely long charged line or a charged conducting cylinder Near line or outside cylinder at r > R E is radial. Choose radial integration path i  f Above is negative for r f > r i with positive Inside conducting cylinder at r < R E is radial. Choose radial integration path i  f Potential inside is constant and equals surface value

Copyright R. Janow Fall 2015 Example 23.12: Potential at a symmetry point near a finite line of charge Uniform linear charge density Charge in length dy Potential of point charge Standard integral from tables: Limiting cases: Point charge formula for x >> 2a Example formula for near field limit x << 2a

Copyright R. Janow Fall 2015 Example: Potential on the symmetry axis of a charged disk  R x z  P a dA=ad  da r Q = charge on disk whose radius = R. Uniform surface charge density  = Q/4  R 2 Disc is a set of rings, each of them da wide in radius For one of the rings: Integrate twice: first on azimuthal angle f from 0 to 2p which yields a factor of 2p then on ring radius a from 0 to R Use Anti- derivative: Double integral (note: ) “Near field” (z<<R): disc looks like infinite non-conducting sheet of charge “Far field” (z>>R): disc looks like point charge