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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 is defined as the product of displacement d and a parallel applied force F. Work = Fd; Units: 1 J = 1 N m Potential Energy U is defined as the ability to do work by virtue of position or condition. (Joules) Kinetic Energy K is defined as the ability to do work by virtue of motion (velocity). (Also in joules) Signs for Work and Energy Work (Fd) is positive if an applied force F is in the same direction as the displacement d. The force F does positive work. The force mg does negative 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 external force F against the g-field increases the potential energy. If released the field gives work 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 external force F against the E-field increases the potential energy. If released the field gives work 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 E-field does positive work on –q decreasing the potential energy. If released from B nothing happens. Work to Move a Charge ++ + + ++Q++ rb Work to move +q from A to B. B A qE ra Avg. Force: Favg F+ kqQ ra rb kQq Work Fd (ra rb ) 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 . B ra 1 1 Work kQq rb ra A 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 mC charge? The P.E. will be positive at point A, because the field can do + work if q is released. Potential Energy: U kQq U r (9 x 10 9 Nm 2 C2 U = 1.35 mJ A +2 nC 8 cm +Q +6 mC )(+6 x 10-6 C)(+2 x 10-9 C) (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 mC 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 A 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 mC )(+6 x 10-6C)(+2 x 10-9C) (0.12 m) DU = UB – UA = 0.9 mJ – 1.35 mJ 0.900 mJ DU = -0.450 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 mC 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 A 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 mC )(6 x 10-6C)(-2 x 10-9C) (0.12 m) B 0.900 mJ DU = +0.450 mJ A – charge moved away from a + charge gains P.E. 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. Electric Potential: U V ; 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 = -800 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 potential of one volt at a given point means that a charge of one coulomb placed at that point will experience a potential energy of one 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 potential due to a positive charge is positive; The potential due to a negative charge is positive. (Use sign of charge.) Example 4: Find the potential at a distance of 6 cm from a –5 nC charge. P. q = –4 mC r 6 cm -- Q -- Q = -5 nC kQ V r 9 Nm2 9 x 10 Negative V at Point P : C 2 (5 x 10-9C) (0.06 m) VP = -750 V What would be the P.E. of a –4 mC charge placed at this point P? U = qV = (-4 x 10-6 mC)(-750 V); U = 3.00 mJ Since P.E. is positive, E will do + work if q is released. Potential For Multiple Charges The Electric Potential V in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge. Q1 - r1 r3 Q3 - A r2 + Q2 kQ1 kQ2 kQ3 VA + + r1 r2 r3 kQ V r Potential is + or – based on sign of the charges Q. 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 10-9C) (0.02 m) VA = 450 V – 2250 V; 2250 V Q1 + +3 nC 6 cm A 2 cm VA = -1800 V 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 10-9C) (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 10-9C) (0.10 m) B 450 V A 2 cm VB = +900 V Q2 = -5 nC Example 5 (Cont.): Discuss meaning of the potentials just found for points A and B. Consider Point A: VA = -1800 V For every coulomb of positive charge placed at point A, the potential energy will be –1800 J. (Negative P.E.) B 2 cm Q1 + +3 nC 6 cm The field holds on to this positive charge. An external force must do +1800 J of work to remove each coulomb of + charge to infinity. A 2 cm Q2 = -5 nC Example 5 (Cont.): Discuss meaning of the potentials just found for points A and B. Consider Point B: VB = +900 V B 2 cm For every coulomb of positive charge placed at point B, the potential energy will be +900 J. (Positive P.E.) For every coulomb of positive charge, the field E will do 900 J of positive work in removing it to infinity. Q1 + +3 nC 6 cm A 2 cm Q2 = -5 nC Potential Difference The potential difference between two points A and B is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential. Potential Difference: VAB = VA - VB WorkAB = q(VA – VB) Work BY E-field The positive and negative signs of the charges may be used mathematically to give appropriate signs. Example 6: What is the potential difference between points A and B. What work is done by the E-field if a +2 mC charge is moved from A to B? B 2 cm VA = -1800 V VB = +900 V Q1 + +3 nC VAB= VA – VB = -1800 V – 900 V VAB = -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 = -5.40 mJ E-field does negative work. Thus, an external force was required to move the charge. Example 6 (Cont.): Now suppose the +2 mC charge is moved from back from B to A? B 2 cm VA = -1800 V VB = +900 V Q1 + +3 nC VBA= VB – VA = 900 V – (-1800 V) VBA = +2700 V 6 cm A 2 cm This path is from high to low potential. Q2 - -5 nC WorkBA = q(VB – VA) = (2 x 10-6 C )(+2700 V) Work = +5.40 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 E field: F = qE +q F = qE VB - - - - Work = Fd = (qE)d Also, Work = q(VA – VB) So that: qVAB = qEd and VAB = Ed The potential difference between two oppositely charged parallel plates is the product of E and d. 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 E-field expressed in volts per meter (V/m) is known as the potential gradient and is equivalent to the N/C. The volt per meter is the better unit for current electricity, the N/C is better electrostatics. Summary of Formulas Electric Potential Energy and Potential Electric Potential Near Multiple charges: WorkAB = q(VA – VB) Oppositely Charged Parallel Plates: kQq U U ; V r q kQ V r Work BY E-field V Ed ; V E d CONCLUSION: Chapter 25 Electric Potential