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Transcript
Electric Potential A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007 Objectives: After completing this module, you should be able to: • Understand an apply the concepts of electric potential energy, electric potential, and electric potential difference. • Calculate the work required to move a known charge from one point to another in an electric field created by point charges. • Write and apply relationships between the electric field, potential difference, and plate separation for parallel plates of equal and opposite charge. Review: Work and Energy Work isis defined defined as as the the product product of of displacement displacement dd Work and aa parallel parallel applied applied force force F. F. and Work == Fd; Fd; Units: Units: 11 JJ == 11 N Nm m Work Potential Energy Energy U U isis defined defined as as the the ability ability to to do do Potential work by by virtue virtue of of position position or or condition. condition. (Joules) (Joules) work Kinetic Energy Energy K K isis defined defined as as the the ability ability to to do do work work Kinetic by virtue virtue of of motion motion (velocity). (velocity). (Also (Also in in joules) joules) by Signs for Work and Energy Work (Fd) (Fd) isis positive positive ifif an an applied applied force forceFF isis Work in the the same same direction direction as as the the displacement displacementd. d. in The force F does positive work. The force mg does negative work. work B F m mg A d The P.E. at B relative to A is positive because the field can do positive work if m is released. P.E. at A relative to B is negative; outside force needed to move m. Gravitational Work and Energy Consider work against g to move m from A to B, a vertical height h. Work = Fh = mgh At level B, the potential energy U is: U = mgh (gravitational) B F m mg A h g The external force does positive work; the gravity g does negative work. The The external external force force FF against against the the g-field g-field increases increases the the potential potential energy. energy. IfIf released released the the field field gives gives work work back. back. Electrical Work and Energy An external force F moves +q from A to B against the field force qE. Work = Fd = (qE)d At level B, the potential energy U is: U = qEd (Electrical) The E-field does negative work; External force does positive work. B ++++ Fe +q + d qE E A - - - - The The external external force force FF against against the the E-field E-field increases increasesthe the potential potential energy. energy. IfIf released released the the field field gives gives work work back. back. Work and Negative Charges Suppose a negative charge –q is moved against E from A to B. Work by E = qEd At A, the potential energy U is: U = qEd (Electrical) No external force is required ! B ++++ qE -q d E A - - - - The The E-field E-field does does positive positive work work on on –q –q decreasing decreasing the the potential potential energy. energy. IfIf released released from from BB nothing nothing happens. happens. Work to Move a Charge ++ + + ++Q++ rb Work to move +q from A to B. qE ra Avg. Force: Favg F kqQ ra rb kQq (ra rb ) Work Fd ra rb kqQ At A: Fa 2 ra kqQ At B: Fb 2 rb Distance: ra - rb 1 1 Work kQq rb ra Absolute Potential Energy ++ + + ++Q++ rb Absolute P.E. is relative to ra 1 1 Work kQq rb ra F It is work to bring +q from infinity to qE point near Q—i.e., from to rb 0 1 1 kQq Work kQq rb rb Absolute Potential Energy: kQq U r Example 1. What is the potential energy if a +2 nC charge moves from to point A, 8 cm away from a +6 C charge? The P.E. will be positive at point A, because the field can do + work if q is released. Potential Energy: U A 8 cm +Q kQq U r (9 x 10 9 Nm 2 C2 U U == 1.35 1.35 mJ mJ +2 nC +6 C )(6 x 10 C)(+2 x 10 C) -6 -9 (0.08 m) Positive potential energy Signs for Potential Energy Consider Points A, B, and C. For +2 nC at A: U = +1.35 mJ Questions: If +2 nC moves from A to B, does field E do + or – work? Does P.E. increase or decrease? A B 12 cm 8 cm C +Q 4 cm +6 C Moving positive q +2 nC The field E does positive work, the P.E. decreases. If +2 nC moves from A to C (closer to +Q), the field E does negative work and P.E. increases. Example 2. What is the change in potential energy if a +2 nC charge moves from to B? Potential Energy: kQq U r A UB (9 x 19 12 cm 8 cm From Ex-1: UA = + 1.35 mJ 9 Nm 2 C2 B +Q +6 C )(6 x 10-6 C)(+2 x 10-9 C) (0.12 m) U = UB – UA = 0.9 mJ – 1.35 mJ 0.900 mJ U U == -0.450 -0.450 mJ mJ Note that P.E. has decreased as work is done by E. Moving a Negative Charge Consider Points A, B, and C. Suppose a negative -q is moved. Questions: If -q moves from A to B, does field E do + or – work? Does P.E. increase or decrease? A B 12 cm 8 cm C +Q 4 cm +6 C Moving negative q - The field E does negative work, the P.E. increases. What happens if we move a –2 nC charge from A to B instead of a +2 nC charge. This example follows . . . Example 3. What is the change in potential energy if a -2 nC charge moves from to B? Potential Energy: kQq U r From Ex-1: UA = -1.35 mJ (Negative due to – charge) UB (9 x 19 9 Nm 2 C2 A UB – UA = -0.9 mJ – (-1.35 mJ) 12 cm 8 cm +Q +6 C )(6 x 10-6 C)(-2 x 10-9 C) (0.12 m) B 0.900 mJ U U == +0.450 +0.450 mJ mJ A from aa ++ charge charge gains gains P.E. P.E. A –– charge charge moved moved away away from Properties of Space Electric Field . E r + ++ + ++Q++ E is a Vector An electric field is a property of space allowing prediction of the force on a charge at that point. F E ; q F qE The field E exist independently of the charge q and is found from: kQ Electric Field : E 2 r Electric Potential Electric potential is another property of space allowing us to predict the P.E. of any charge q at a point. U Electric V ; Potential: q U qV The units are: joules per coulomb (J/C) U P. V q r + ++ + ++Q++ Potential For example, if the potential is 400 J/C at point P, a –2 nC charge at that point would have P.E. : U = qV = (-2 x 10-9C)(400 J/C); U U == -800 -800 nJ nJ The SI Unit of Potential (Volt) From the definition of electric potential as P.E. per unit charge, we see that the unit must be J/C. We redefine this unit as the volt (V). U V ; q 1 joule 1 volt = 1 coulomb A A potential potential of of one one volt volt at at aa given given point point means means that that aa charge charge of of one one coulomb coulomb placed placed at at that that point point will will experience experience aa potential potential energy energy of of one one joule. joule. Calculating Electric Potential Electric Potential Energy and Potential: kQq U U ; V r q Substituting for U, we find V: V kQ V r kQq r q kQ kQ P. V r r + ++ + ++Q++ Potential r The The potential potential due due to to aa positive positive charge charge isis positive; positive; The The potential potential due due to to aa negative negative charge charge isis positive. positive. (Use (Use sign sign of of charge.) charge.) Example 4: Find the potential at a distance of 6 cm from a –5 nC charge. P. q = –4 C r 6 cm -- Q -- Q = -5 nC kQ V r 9 x 10 Negative V at Point P : 9 Nm 2 C 2 (5 x 10-9 C) (0.06 m) VVPP == -750 -750 V V What would be the P.E. of a –4 C charge placed at this point P? U = qV = (-4 x 10-6 C)(-750 V); U U == 3.00 3.00 mJ mJ Since released. Since P.E. P.E. isis positive, positive, EE will will do do ++ work work ifif qq isis released. Potential For Multiple Charges The Electric Electric Potential Potential VV in in the the vicinity vicinity of of aa number number The of charges charges isis equal equal to to the the algebraic algebraic sum sum of of the the of potentials due due to to each each charge. charge. potentials Q1 - r1 r3 Q3 - A r2 + Q2 kQ1 kQ2 kQ3 VA r1 r2 r3 kQ V r Potential isis ++ or or –– based based on on sign sign of of the the charges charges Q. Q. Potential Example 5: Two charges Q1= +3 nC and Q2 = -5 nC are separated by 8 cm. Calculate the electric potential at point A. B kQ1 kQ2 VA r1 r2 2 cm 9 x 10 (3 x 10 C) kQ1 C2 450 V r1 (0.06 m) kQ2 r2 9 x 10 9 Nm 2 9 Nm 2 C 2 -9 (5 x 10-9 C) (0.02 m) VA = 450 V – 2250 V; 2250 V Q1 + +3 nC 6 cm A 2 cm V -1800 V V VAA == -1800 Q2 = -5 nC Example 5 (Cont.): Calculate the electric potential at point B for same charges. kQ1 r1 kQ2 r2 kQ1 kQ2 VB r1 r2 9 x 109 Nm 2 C2 (3 x 10-9 C) (0.02 m) 9 x 10 9 Nm 2 C 2 VB = 1350 V – 450 V; 2 cm 1350 V Q1 + +3 nC 6 cm (5 x 10-9 C) (0.10 m) B 450 V A 2 cm V VBB == +900 +900 V V Q2 = -5 nC Example 5 (Cont.): Discuss meaning of the potentials just found for points A and B. Consider Point A: V VAA == -1800 -1800 V V For For every every coulomb coulomb of of positive positive charge charge placed placed at at point point A, A, the the potential potential energy energy will will be be –1800 –1800 J.J. (Negative (Negative P.E.) P.E.) The The field field holds holds on on to to this this positive positive charge. charge. An An external external force force must must do do +1800 +1800 JJ of of work work to to remove remove each each coulomb coulomb of of ++ charge charge to to infinity. infinity. B 2 cm Q1 + +3 nC 6 cm A 2 cm Q2 = -5 nC Example 5 (Cont.): Discuss meaning of the potentials just found for points A and B. Consider Point B: V +900 V V VBB == +900 For For every every coulomb coulomb of of positive positive charge charge placed placed at at point point B, B, the the potential potential energy energy will will be be +900 +900 J.J. (Positive (Positive P.E.) P.E.) For For every every coulomb coulomb of of positive positive charge, charge, the the field field EE will will do do 900 900 JJ of of positive positive work work in in removing removing itit to to infinity. infinity. B 2 cm Q1 + +3 nC 6 cm A 2 cm Q2 = -5 nC Potential Difference The The potential potential difference difference between between two two points points A A and and BB isis the the work work per per unit unit positive positive charge charge done done by by electric electric forces forces in in moving moving aa small small test test charge charge from from the the point point of of higher higher potential potential to to the the point point of of lower lower potential. potential. Potential =V -V Potential Difference: Difference: VVAB AB = VAA - VBB Work = q(V – V ) WorkAB AB = q(VAA – VBB) Work Work BY BY E-field E-field The The positive positive and and negative negative signs signs of of the the charges charges may may be be used used mathematically mathematically to to give give appropriate appropriate signs. signs. Example 6: What is the potential difference between points A and B. What work is done by the E-field if a +2 C charge is moved from A to B? B 2 cm V Q1 + +3 nC VBB == +900 VAA == -1800 -1800 V V V +900 V V VAB= VA – VB = -1800 V – 900 V VVAB = -2700 V AB = -2700 V Note point B is at higher potential. 6 cm A Q2 - 2 cm -5 nC WorkAB = q(VA – VB) = (2 x 10-6 C )(-2700 V) Work Work == -5.40 -5.40 mJ mJ E-field does negative work. Thus, an external force was required to move the charge. Example 6 (Cont.): Now suppose the +2 C charge is moved from back from B to A? B 2 cm V Q1 + +3 nC VBB == +900 VAA == -1800 -1800 V V V +900 V V VBA= VB – VA = 900 V – (-1800 V) 6 cm A 2 cm This path is from VVBA = +2700 V BA = +2700 V high to low potential. Q2 - -5 nC WorkBA = q(VB – VA) = (2 x 10-6 C )(+2700 V) Work Work == +5.40 +5.40 mJ mJ E-field does positive work. The work is done BY the E-field this time ! Parallel Plates Consider Two parallel plates of equal VA + + + + and opposite charge, a distance d apart. Constant Constant EE field: field: FF == qE qE Work = Fd = (qE)d +q F = qE VB - - - - Also, Work = q(VA – VB) So that: qVAB = qEd and VVAB = Ed AB = Ed The The potential potential difference difference between between two two oppositely oppositely charged charged parallel parallel plates plates isis the the product product of ofEE and and dd.. E Example 7: The potential difference between two parallel plates is 800 V. If their separation is 3 mm, what is the field E? VA + + + + +q F = qE VB - - - - V Ed ; E V E d 80 V E 26, 700 V/m 0.003 m The The E-field E-field expressed expressed in in volts volts per per meter meter (V/m) (V/m) isis known known as as the the potential potential gradient gradient and and isis equivalent equivalent to to the the N/C. N/C. The The volt volt per per meter meter isis the the better better unit unit for for current current electricity, electricity, the the N/C N/C isis better better electrostatics. electrostatics. Summary of Formulas Electric Electric Potential Potential Energy Energy and and Potential Potential kQq U U ; V r q Electric Electric Potential Potential Near Near Multiple Multiple charges: charges: kQ V r Work = q(V – V ) WorkAB AB = q(VAA – VBB) Work Work BY BY E-field E-field Oppositely Oppositely Charged Charged Parallel Parallel Plates: Plates: V Ed ; V E d CONCLUSION: Chapter 25 Electric Potential