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Electric Potential Energy versus Electric Potential Calculating the Potential from the Field

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

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Electric Potential Energy versus Electric Potential Calculating the Potential from the Field

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  1. 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

  2. Q1f = ?? Q2f = ?? wire Open valve, water flows. What determines final water levels? Mechanical analogy: Water pressure Potential energy of water molecule at the top surface = mgy P2 = rgy2 P1 = rgy1 PE/unit mass = gy Pressure = rgy X Electrostatics: Two metal spheres, different radii, one charged Initially Connect wire between spheres, then disconnect it Q10= 10 C r1= 10 cm • Are final charges equal? • What determines how the charge redistributes itself? • What if spheres are identical? Q20= 0 C r2= 20 cm

  3. ELECTROSTATIC POTENTIAL Definition: Electrostatic Potential equals Electrostatic Potential Energy per unit (test) charge Potential Energy DU: • DU = Work done ~ force x displacement or charge x field x displacement. Potential DV: • DV = - Work done on a test charge / test charge ~ - field x displacement. = Change in PE of a test charge as it is displaced, per unit test charge. Units, Dimensions: • Potential Energy DU: Joules • Potential DV: [DU] / [q] are Joules/C. = VOLTS • E field units: [V]/[d] are Volts / meter – same as N/C. Both DU and DV: • use a reference level (possibly at infinite distance) & are functions of position. • represent conservative forces via scalar fields (does gravity). • can describe the motion of objects in conjunction the Work-KE theorem &/or mechanical energy conservation, as an alternative to using the Second Law (F = qE = ma).

  4. Ds I II Force may vary in direction and magnitude along the path: III IV Result of a “Path Integration” in general can depend on integration path taken Reminder: Work Done by a Constant Force 5-1: In each example shown in the sketch, a force F has the same magnitude but the direction varies. The displacements of the object are all to the right with 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, IV, III, II • II, I, IV, III • III, II, IV, I • I, IV, II, III • III, IV, I, II

  5. Definitions: Electrostatic Potential versus Potential Energy POTENTIAL ENERGY DIFFERENCE: Charge q0 moves from i to f along ANY path path integral POTENTIAL DIFFERENCE: (from basic definition) Potential is work done by field per unit charge ( Evaluate integral on ANYpath from i to f ) E field is “conservative”, like gravity • Work dW done BY THE FIELD on a test charge moving from i to f does notdepend on path taken. • Work done changes potential energy and potential. • Work done around any closed path equals zero.

  6. i E o f Dx CHOOSE THE SIMPLE PATH SHOWN 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 are also volts/meter To convert potential to PE just multiply by q0 What are signs of DU and DV if test charge is negative? EXAMPLE: Find change in potential DV as test charge +q0 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

  7. f i E Work and PE : Who/what does positive or negative work? 5-2: In the figure, we exert an external force that moves a proton from point i to point f in a uniform electric field as shown. Which of the following statements is true? 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. Hint: which directions pertain to displacement and force?

  8. Summary and details • Reference levels can be: • Relative: Choose arbitrary zero reference level for ΔU or ΔV. …or… • Absolute: Set Ui = 0 for all charges that started 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 • New name for electric field units: • 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 = (1.60×10-19 C)(1 J/C) = 1.60×10-19 J • The field is created by a charge distribution somewhere else. • A test charge q0 moved between i and f gains or loses potential energy DU. • DU and DV do not depend on path. DV also does not depend on q0 (test charge). • Use Work-KE theorem to link potential differences to motion.

  9. Example: Potential Function for a Point Charge Find potential V(R) a distance R from a point charge q : Positive for q > 0, Negative for q < 0 Inversely proportional to r1 NOT r2 For potential ENERGY: same method but integrate force Shared PE between q and q0 Overall sign depends on both signs • Reference level: V(r) = 0 for r = infinity (makes sense) • Field is conservative  choose radial integration path for E(r) • DU = q0DV = work done on a test charge as it moves to final location without change in kinetic energy

  10. Vi > Vf DV= 0 DV= 0 DV= 0 E = -Gradient of V = -spatial rate of change of V Vfi Electric field lines are perpendicular to the equipotentials - zero work is done moving charge along an equipotential q and On an “Equipotential Surface” (Contour) voltage and potential energy are constant; i.e. DV=0, DU=0 for Dsnormal to E Steepest Change

  11. Chapter 23 Charged Conductor with Teardrop Shape Where on the surface is the field strongest? CONDUCTORS ARE ALWAYS EQUIPOTENTIALS - Charge on conductors moves to make Einside = 0 - Charge is on the outer surface - Esurf is perpendicular to surface = s/e0 so DV = 0 along any path in or on a conductor DemonstrationSource: Pearson Study Area - VTD https://mediaplayer.pearsoncmg.com/assets/secs-vtd34_chargeovoid Discussion: • Blunt and sharp ends are at the same potential • How do the potential, field, and surface charge densities differ for the blunt and sharp ends? • E = s / e0 just outside the surface and is outward (Gauss’ Law) • At the sharp end q = sA is larger, so s is larger and so is E

  12. Examples of Equipotential Surfaces Point charge or outside a sphere of charge Uniform Field Dipole Field Equipotentials are planes (evenly spaced) Equipotentials are spheres (spaced for 1/r) Equipotentials are not simple shapes

  13. s+ s- L Dx E is a known, constant function, so Vi Vf E = 0 Assume: • s = 1 nanoCoulomb/m2 • Dx = 1 cm & points from negative to positive plate • Uniform field E (from + to – plate) A positive test charge +q gains potential energy DU = qDV as it moves from - plate to + plate along any path (including an external circuit) Example: Find potential difference between oppositely charged plates of a parallel plate capacitor • Infinite sheets of uniform charge (Dx << L) • Equal and opposite surface charges • All charge moves to inner surfaces (opposite charges attract)

  14. Connect wire between spheres – charge moves Conducting spheres come to same potential Net charge on system stays constant, but redistributes to make the system an equipotential. Spherical symmetry. r1= 10 cm V10= 90,000 V. wire r2= 20 cm V20= 0 V. Q20= 0 V. Initially (use shell theorem): Find the final charges: Find the final potential(s): Conductors are always equipotentials Example: Two spheres, different radii, one charged to 90,000 V.

  15. d c R a b Definition: Apply Gauss’ Law with GS just inside shell: Potential is continuous across conductor surface – field is not E(r) V(r) Eoutside= kq / r2 Einside=0 Voutside= kq / r Vinside=Vsurf r R r R Example: Find potential Va at point “a” inside a hollow conducting shell Assume Vb = 18,000 Volts on surface. Vc = Vb (conducting shell is an equipotential). Shell can be any closed surface (sphere or not)

  16. Comparison: For the electric field… Unit vector from ri to r Potential due to a group of point charges • Use superposition for n point charges • Potential is a scalar… so … the sum is an algebraic sum, not a vector sum. E may be zero where V does not equal to zero. V may be zero where E does not equal to zero.

  17. Examples: potential due to point charges TWO EQUAL CHARGES – Point P at the midpoint between them +q +q +q -q P 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. DIPOLE – Otherwise positioned as above d Use Superposition Note: E may be zero where V does not = 0 V may be zero where E does not = 0

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

  19. q -q q q A B q q q -q q q -q q q -q C D E Electric Field and Electric Potential 5-3: Which of the following figures have both V=0 and E=0 at the red point?

  20. Gauss Law result: Field outside cylinder at r > R Field inside = 0 for r < R E is radial. Choose radial integration path if Above is negative for rf > ri with l positive inside conducting cylinder for r < R Take E to be radial, integrate over radial path if Potential inside is constant and equals surface value Finding potential if E is known: use the definition of V(Example 23.10: Potential near an infinitely long, charged, conducting cylinder mimics a line of charge for points outside the cylinder)

  21. 2. Find an expression for dq, the charge in a “small” chunk of the distribution, in terms of l, s, or r Typical challenge: express above in terms of chosen coordinates 3. At point P, dV is the differential contribution to the potential due to a point-like chargedq located in the distribution. Use symmetry to simplify. 4. Use “superposition”. Add up (integrate) the contributions over the whole charge distribution, varying the displacement r as needed. Scalar VP. 5. Field E can be gotten from potential by taking the “gradient”: Rate of potential change perpendicular to equipotential Method for finding potential function V due to a continuous charge distribution: E not known 1. Assume V = 0 infinitely far away from charge distribution (finite size)

  22. z P All scalars - no need to worry about direction r z Looks like a point charge formula, but r is on the ring y a f dq As a  0 or z  inf, V  point charge As z  0, V  kQ/a x 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 Example 23.11: Potential along Z-axis of a ring of charge Q = charge on the ring l = uniform linear charge density = Q/2pa r = [a2 + z2]1/2 = distance from dq to “P” ds = arc length = adf

  23. Supplementary Material

  24. For q negative V is negative (funnel) r V(r) 1/r r Visualizing the potential function V(r) for a positive point charge (2 Dimensions)

  25. Gradient is measured along the path. For the case at left it would be: 15 • The gradient of the potential energy is the gravitational force component along path Dl: 15% 100 Dl Dh Dx q Distinction: Slope, Grade, Gradients in Gravitational Field EXTRA TOPIC Height contours portray constant gravitational potential energy DU = mgDh. Force is along the gradient, perpendicular to a potential energy contour. • Grade means the same thing as slope. A 15% grade is a slope of • The GRADIENT of height (or gravitational potential energy) is a vector field representing steepness (or gravitational force) • In General: The GRADIENTS of scalar fields are vector fields.

  26. equipotentials q E E = -spatial rate of change of V = -Gradient of V The field E(r) is the gradient of the potential EXTRA TOPIC Uniform field, normal to equipotential surfaces Component of ds on E changes potential Component of ds normal to E produces no change For a path along equipotential, DV = 0

  27. Uniform linear charge density Charge in length dy Potential of point charge Use infinite series approximation From standard integral tables: Limiting cases: • Point charge formula for x >> 2a • Example 23.10 formula for near field limit x << 2a Example 23.12: Potential at symmetry point P near a finite line of charge

  28. Start with The following will be used below to integrate: • Integrate over the charge distribution using interval 0 < x < L • Result Optional Example: Potential Due to a Charged Rod • 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.

  29. P dA=adfda q r z R a f Double integral 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 x Use Anti-derivative: Homework Example: Potential on the symmetry axis of a charged disk Q = charge on disk whose radius = R. Uniform surface charge density s = Q/pR2 Disc is a set of rings, radius a, each of them da wide in radius For one of the rings:

  30. P dA=adfda q r z R An approximation using infinite series a f x Limiting cases: Potential on the symmetry axis of a charged disk “Far field”: assume z>>R Disc mimics a point charge far away “Near field”: assume z<<R Disk mimics an infinite non-conducting sheet of charge

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