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Electric Potential

Electric Potential. A. GPE = mgΔh. GPE = mgh A – mgh B. F = mg. GPE = Work (W) required to raise or lower the book. - Where W = (F gravity )( Δh). B. h A. h B. Gravitational Potential Energy. + + + + + + + + +. ΔEPE = q o EΔd.

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Electric Potential

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  1. Electric Potential

  2. A GPE = mgΔh GPE = mghA – mghB F = mg • GPE = Work (W) required to raise or lower the book. • -Where W = (Fgravity)(Δh) B hA hB Gravitational Potential Energy

  3. + + + + + + + + + ΔEPE = qoEΔd ΔEPE = qoEdA – qoEdB dA -WE(AB) = qoEdA – qoEdB A B + + -WE(AB) = FedA – FedB dB Fe = qoE Fe = qoE - - - - - - - - - - Electric Potential Energy • Does a proton at rest at point A have more or less potential energy than it would at point B? More

  4. -qo F = kqqo Ue = Fr = kqqo E r r2 r +q Electric Potential Energy of Point Charges • Much like the book is attracted to the earth due to gravity, two unlike charges are attracted to one another. • Conversely, like charges repel. • It takes positive work to move unlike charges away from one another and negative work to move them closer together. EPE

  5. -qo To change the energy level from UA to UB, it requires positive work (W). -W = UB – UA -W = kqqo – kqqo rB rA rB -qo rA +q +q B A Electric Potential Energy and Work of Point Charges

  6. -qo +q Electric Potential Energy • What would happen if the charged particle q was fixed in place and then particle qo was suddenly released from rest? • It would accelerate away from q. • It would accelerate towards q. • It would stay where it is. • How would the potential energy of this system change? • It would increase. • It would decrease. • It would remain the same.

  7. Electric Potential SI Units: joule/coulomb = 1 volt (V) • The Electric Potential Difference is equal to the Work required to move a test charge from infinity to a point in an electric field divided by the magnitude of the test charge. • The Electric Potential is the energy per unit of charge (J/C). Point Charges only

  8. Example 1: Electric Potential • An object with 2.5C of charge requires 1.00x10-3 Joules of energy to move it through an electric field. What is the potential difference through which the charge is moved?

  9. -WAB kq kq qo rB rA VB - VA = = - Relationship Between Electric Potential and Distance(point charges) • Consider relationship between V and r. • What happens to V as rB goes to ? • As r increases, i.e., as rB, V  0. • The relationship above reduces to: V = kq/r • The sign of the charge will determine if the electric potential is positive or negative. • When two or more charges are present, the total electric potential is the sum total from all the charges present in the system.

  10. Q1 V = + + Q2 r1 r2 Q3 r3 P Electric Potential(point charges) • Consider the following system of three point charges. What is the electric potential that these charges give rise to at some arbitrary point P? • Use superposition to determine V. kQ1 kQ2 kQ3 r1 r2 r3 • Note that the electric potential can be determined from any arbitrary point in space.

  11. Q1 Q2 r1 r2 Q3 r3 qo Electric Potential and Electrical Potential Energy/Work (point charges) • If we now move a test charge from infinity to point P, we can determine the potential energy of the system or the work required to the test charge to its new location. • Remember: work = energy.

  12. +3.00 μC -6.10 μC B A 0.5 m 0.5 m Example 2: Two Point Charges Two point charges, +3.00 µC and -6.10 µC, are separated by 1.00 m. What is the electric potential midway between them? Vtotal = VA + VB = kqA/rA + kqB/rB VA = (8.99 x 109 Nm2/C2)(-6.10 x 10-6C) = -109678 V 0.5m VB = (8.99 x 109 Nm2/C2)(3.00 x 10-6C) = 53,940 V 0.5m Vtotal = -55700 V

  13. + + + + + + + + + + + ++++ - - - - - - - - - - - ---- qo B Uniform Electric Field Two equal and oppositely charged plates qo C qo A Characteristics of a Capacitor E • Since the electric field is constant, the force acting on a charged particle will be the same everywhere between the plates. • Fe = qoE FA = FB = FC

  14. + + + + + + + + + + + ++++ - - - - - - - - - - - ---- B F = qoE dB dA qo Electric Potential and Work in a Capacitor D WAB = F·dB - F·dA A qo WAB = qoEd F = qoE (Ue) -WAB qo qo V = = qo C If WAB = qoEd, then what is WCD? • WCD = 0 Joules because the force acts perpendicular to the direction of motion. • Do you remember that W = F·d·cos?

  15. + + + + + + + + + + + ++++ - - - - - - - - - - - ---- d Electric Potential of a Capacitor – An alternative • From mechanics, W = Fd. • From the previous slide, W = qoEd • From the reference table, V = W/qo Two equal and oppositely charged plates A B qo F = qoE Uniform Electric Field V = WAB/qo = Fd/qo = qoEd/qo= Ed

  16. d Example 3:Parallel Plates A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.50 mm. When an electric spark jumps between them, the magnitude of the electric field is 4.8 x 107 V/m. What is the magnitude of the potential difference V between the conductors? V = Ed V = (4.8 x 107 V/m)(5.0 x 10-4m) V = 24,000V

  17. Example 4: Parallel Plates A proton and an electron are released from rest from a similarly charged plate of a capacitor. The electric potential is 100,000 V and the distance between the two plates is 0.10 mm. • Which charge will have greater kinetic energy at the moment it reaches the opposite plate? • Determine the amount of work done on each particle. • Determine the speed of each particle at the moment it reaches the opposite plate. • Determine the magnitude of the force acting on each particle. • Determine the magnitude of the acceleration of each particle.

  18. + + + + + + + + + + + ++++ - - - - - - - - - - - ---- d Example 4: Parallel Plates(cont.) • Begin by drawing a picture and listing what is known: • V = 100,000V • d = 0.10 mm = 1.0 x 10-4m • qe = qp = 1.6 x 10-19C (ignore the sign. We are only interested in magnitude.) p+ e-

  19. Example 4: Parallel Plates(#1 & #2) • For #1, you could answer #2 first to verify. • The answer is that the kinetic energy of both particles will be the same • Why? • because of the formula needed in question #2 applies to both charges, and work = energy. • Hence: Wproton = Welectron qprotonV = qelectronV Wproton = Welectron = (1.6x10-19C)(100,000V) Wproton = Welectron = 1.6x10-14 J

  20. Example 4: Parallel Plates(#3) • Apply the work-energy theorem to determine the final speed of the electron and proton. W = KE • Since the initial kinetic energy is equal to 0J: W = KEf W = ½ mvf2 • Proton: • Electron:

  21. Example 4: Parallel Plates(#4) • Since F = qE, it will be the same for both particles because their charges are the same and the electric field is uniform between two parallel plates. • We also know that W = Fd. Since we know the distance between the plates and the work done to move either charge from one plate to another, we can determine the force as follows:

  22. Example 4: Parallel Plates(#5) • Since we have the force acting on each particle, we can now calculate the acceleration of each particle using Newton’s 2nd Law.

  23. Equipotential Lines • Equipotential lines denote where the electric potential is the same in an electric field. • The potential is the same anywhere on an equipotential surface a distance r from a point charge, or d from a plate. • No work is done to move a charge along an equipotential surface. Hence VB = VA (The electric potential difference does not depend on the path taken from A to B). • Electric field lines and equipotential lines cross at right angles and point in the direction of decreasing potential.

  24. + + + + + + + + + + + ++++ - - - - - - - - - - - ---- Lines of Equipotential Note: Electric field lines and lines of equipotential intersect at right angles. Equipotential Lines • Parallel Plate Capacitor Electric Field Lines Decreasing Electric Potential / Voltage

  25. Note: Electric field lines and lines of equipotential intersect at right angles. Lines of Equipotential + Equipotential Lines • Point Charge Electric Field Lines Note: A charged surface is also an equipotential surface! Decreasing Electric Potential / Voltage

  26. Equipotential Lines (Examples) • http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html

  27. Key Ideas • Electric potential energy (U) is the work required to bring a positive unit charge from infinity to a point in an electric field. • Electric potential (V) is the change in energy per unit charge as the charge is brought from one point to another. • The electric field between two charged plates is constant meaning that the force is constant between them as well. • The electric potential between two points is not dependent on the path taken to get there. • Electric field lines and lines of equipotential intersect at right angles.

  28. + + + + + + + + + + + ++++ - - - - - - - - - - - ---- B F = qoE dB dA qo Electric Potential Energy and Work in a Uniform Electric Field A qo F = qoE Note: The force acting on the charge is constant as it moves from one plate to another because the electric field is uniform. WAB = EPEB – EPEA WAB = FdB – FdA WAB = qoEdB – qoEdA WAB = qoE(dB – dA) = qoEd

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