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Topic 6.2 Extended D – Electric potential energy and p.d.

Topic 6.2 Extended D – Electric potential energy and p.d. Topic 6.2 Extended D – Electric potential energy and p.d. G RAVITATIONAL P OTENTIAL E NERGY D IFFE RENCE - P OINT M ASS.

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Topic 6.2 Extended D – Electric potential energy and p.d.

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  1. Topic 6.2 ExtendedD – Electric potential energy and p.d.

  2. Topic 6.2 ExtendedD – Electric potential energy and p.d. GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE - POINT MASS In mechanics we began with vectors, and then found easier ways to solve problems - namely energy considerations. rf In the last chapter we looked at electric vectors. Now we want to look at electric energy. This will simplify finding solutions to electrical problems. r0 Consider a mass in a gravitational field: If we raise the mass from a height of r0 to a height of rfand release it, we know it will relinquish the potential energy we gave to it - converting it to kinetic energy.

  3. Gravitational Potential Energy Difference (Point Mass) Topic 6.2 ExtendedD – Electric potential energy and p.d. GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE - POINT MASS The gravitational force is given by GmM r2 rf FG = and the work done against gravity is the potential energy difference UG = FG rf- FG r0 GmM r02 GmM rf2 r0 •r0 •rf - = GmM r0 GmM rf UG = -

  4. A q0 q0 Electric Potential Energy Difference (Point Charge) Topic 6.2 ExtendedD – Electric potential energy and p.d. ELECTRIC POTENTIAL ENERGY DIFFERENCE - POINT CHARGE An electric field has the same properties as a gravitational field. If we place a test charge q0 in an electric field we must do work on it in moving it from one place to another. If we release the test charge it will relinquish the potential energy we gave to it, just as the mass did in the gravitational field. r0 The potential energy difference of the test charge is given by Ue = Fe rf- Fe r0 kq0Q r02 kq0Q rf2 •r0 •rf - = kq0Q r0 kq0Q rf Ue = - rf

  5. Gravitational Potential Difference (Point Mass) Electric Potential Difference (Point Charge) Note the missing word "Energy" Note the missing word "Energy" Note the missing word "Energy" Topic 6.2 ExtendedD – Electric potential energy and p.d. ELECTRIC POTENTIAL DIFFERENCE - POINT CHARGE We define the electricpotential differenceVto be the potential energy per unit charge. Thus kQ r0 Ue q0 kQ rf - = V = We eliminate the need for the presence of a test charge in the expression for electric potential energy. We essentially have a quantity that tells us the potential energy per unit positive charge. We can also define the gravitationalpotential differenceVgto be the potential energy per unit mass. Thus GM r0 UG m GM rf - = VG =

  6. Gravitational Potential Energy Difference Gravitational Potential Difference Topic 6.2 ExtendedD – Electric potential energy and p.d. GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE - LOCAL Of course, at the surface of the earth we have a local gravitational field, represented with a bunch of g's: Consider a baseball in a gravitational field: If we raise the ball from a height of h0 to a height of hfand release it, we know it will relinquish the potential energy we gave to it - converting it to kinetic energy. hf The potential energy of the ball in the local gravitational field is given by h UG = FG hf - FG h0 = mghf - mgh0 h0 GRAVITATIONAL POTENTIAL DIFFERENCE - LOCAL UG m W m gh = VG = =

  7. + + + + + + + + + + + + + + + + + - - - - - - - - - - - - - - - - - FYI: The E-field will be perpendicular to the plates. Why? FYI: The E-field will act just like the local gravitational field. Topic 6.2 ExtendedD – Electric potential energy and p.d. ELECTRIC FIELD IN A PARALLEL PLATE CAPACITOR A parallel plate capacitor is essentially two metal plates separated by a small distance. If we connect the two plates to a battery, electrons from the negative side of the battery will be "loaded" onto one plate, and electrons from the other plate will be drawn off of the other plate into the positive side of the battery. - + -

  8. Electric Potential Energy Difference (Parallel Plates) Electric Potential Difference (Parallel Plates) Topic 6.2 ExtendedD – Electric potential energy and p.d. ELECTRIC POTENTIAL ENERGY DIFFERENCE - PLATES The potential energy of a test charge in the local electric field is given by UE = FE df - FE d0 = q0Edf - q0Ed0 ELECTRIC POTENTIAL DIFFERENCE - PARALLEL PLATES U q0 W q0 Ed = V = =

  9. FYI: Perhaps you recall that only CONSERVATIVE forces can have associated potential energies. The electric force, like the gravitational force is, in fact, conservative. This means that the potential energy difference does NOT depend on the path of the charge. Topic 6.2 ExtendedD – Electric potential energy and p.d. All right. Let’s do some practice problems. An electron in the vicinity of a proton is moved from a distance of 1 meter to a distance of 2 meters. What is the potential energy difference between initial and final positions? Note first that we seek the potential ENERGY difference. kq0Q r0 kq0Q rf Ue = (use the point charge form) - 1 rf 1 r0 Ue = kq0Q - 1 2 1 1 - Ue = (9×109)(-1.6×10-19)(1.6×10-19) Ue = +1.152×10-28 J FYI: Note that the electron-proton system gained energy. FYI: If the electron had begun at 2 m and ended at 1 m, our change in potential energy would have been negative. Why?

  10. Topic 6.2 ExtendedD – Electric potential energy and p.d. An electron in the vicinity of a proton is moved from a distance of 1 meter to a distance of 2 meters. What is the potential difference between initial and final positions? Potential difference is just potential ENERGY difference divided by q0– in this case the charge of the electron: +1.152×10-28 J -1.6×10-19 C Ue q0 -7.2×10-10 V = V = = Note that the SI unit for electric potential difference is the joule / coulomb or volt. FYI: These are the same volts that you are familiar with. FYI: We could have bypassed the previous problem and found V directly by using kQ r0 Ue q0 kQ rf - = V =

  11. Topic 6.2 ExtendedD – Electric potential energy and p.d. An electron is located next to the negative plate of a parallel plate pair, and released from rest. The positive plate is located 2 cm away and the electric field strength between the plates is 250 N/C. (a) Find the potential energy lost by the electron as it travels to the positive plate. Again, choose the correct formula – that for ENERGY: (use the parallel plate form) UE = q0Edf - q0Ed0 UE = q0E(df - d0) UE = (-1.6×10-19)(250)(0.02 - 0) UE = -8×10-19 J

  12. Topic 6.2 ExtendedD – Electric potential energy and p.d. An electron is located next to the negative plate of a parallel plate pair, and released from rest. The positive plate is located 2 cm away and the electric field strength between the plates is 250 N/C. (b) Find the speed of the electron as it strikes the positive plate. Use energy considerations. K + UE = 0 K = -UE K = 8×10-19 J 1 2 mv2 = 8×10-19 1 2 (9.11×10-31)v2 = 8×10-19 v= 1.3×106 m/s FYI: Don’t confuse voltage V with velocity v.

  13. Topic 6.2 ExtendedD – Electric potential energy and p.d. An electron is located next to the negative plate of a parallel plate pair, and released from rest. The positive plate is located 2 cm away and the electric field strength between the plates is 250 N/C. (c) Find potential difference between the two plates. Pick the right formula: (use the parallel plate form) V = Ed V = (250)(0.02) V = 5 V

  14. FYI: The NEGATIVE sign means that these charges have a lower potential energy than they would if they were farther away. The highest potential energy for these particular charges would be ZERO, at . Topic 6.2 ExtendedD – Electric potential energy and p.d. FYI: If all the charges were POSITIVE (or NEGATIVE), the total potential energy would be positive. The lowest potential energy would occur at , and would be ZERO. Find the total electrostatic potential energy in the charge configuration shown. Assume the charges have been assembled from infinity! q3 = -2C 0.25 m 0.25 m q1 = +5C q2 = +1C 0.25 m Pick the right formula and work in pairs: 0 kqaqb  kqaqb rf Ue = (use the point charge form) - kq1q3 r13 kq1q2 r12 kq2q3 r23 U13 = U12 = U23 = -k12 0.25 -k52 0.25 k51 0.25 U23 = U13 = U12 = U23 = -0.072 J U13 = -0.36 J U12 = 0.18 J = -0.252 J Utotal = 0.18 J + -0.36 J + -0.072 J

  15. Question: How would you find their speeds after they have separated to infinity? Find their speed now. Is it what you expected? Topic 6.2 ExtendedD – Electric potential energy and p.d. Two electrons are placed 1 cm apart and released. At what speed are they each traveling after their separation is 10 cm? Use energy considerations: K + UE = 0 0 Kf + Uf = K0 + U0 kqq rf kqq r0 1 2 1 2 mv2 mv2 + = + 1 r0 1 rf mv2 = kq2 - kq2 m 1 r0 1 rf - v2 = (9×109)(-1.6×10-19)2 9.11×10-31 1 .01 1 .10 - v2 = v = 150.87 m/s

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