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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. GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE In mechanics we began with vectors, - POINT MASS 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. Consider a mass in a gravitational field: If we raise the mass from a height of r0 to a height of rf and release it, we know it will relinquish the potential energy we gave to it converting it to kinetic energy. r0 Topic 6.2 Extended D – Electric potential energy and p.d. GRAVITATIONAL POTENTIAL ENERGY DIFFERENCE The gravitational force is given by - POINT MASS GmM r2 and the work done against gravity is the potential energy difference FG = UG = FG rf - FG r0 GmM GmM •r0 •r = f 2 2 r0 rf UG = GmM rf - GmM r0 Gravitational Potential Energy Difference (Point Mass) rf r0 Topic 6.2 Extended D – 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 A relinquish the potential energy we gave to it, just as the mass did in the q0 r 0 gravitational field. The potential energy difference of the test charge is given by Ue = Fe rf - Fe r0 kq0Q kq0Q •r0 •r = f 2 2 r0 rf Ue = kq0Q rf - kq0Q r0 Electric Potential Energy Difference (Point Charge) rf q0 Note the missing word Topic"Energy" 6.2 Extended D – Electric potential energy and p.d. ELECTRIC POTENTIAL DIFFERENCE - POINT CHARGE We define the electric potential difference V to be the potential energy per unit charge. Thus Ue = V = q0 kQ - kQ r0 rf Electric Potential Difference (Point Charge) 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 gravitational potential difference Vg to be the potential energy per unit mass. Thus VG = UG = m GM - GM r0 rf Gravitational Potential Difference (Point Mass) Topic 6.2 Extended D – 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 hf and release it, we know it will relinquish the potential energy we gave to it converting it to kinetic energy. The potential energy of the ball in the local h gravitational field is given by UG = FG hf - FG h0 = mghf - mgh0 Gravitational Potential Energy Difference GRAVITATIONAL POTENTIAL DIFFERENCE VG = UG = m W m = gh hf - LOCAL Gravitational Potential Difference h0 FYI: The E-field will be perpendicular to the plates. Why? Topic 6.2 Extended FYI: The E-field will act just like the local gravitational field. D – 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. + +- + + + + + + + + + + + + + + + + ---------------- Topic 6.2 Extended D – 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 W U = Ed V = = q0 q0 Electric Potential Energy Difference (Parallel Plates) - PARALLEL PLATES Electric Potential Difference (Parallel Plates) FYI: Perhaps you recall that only CONSERVATIVE forces can have Topic The 6.2 Extended associated potential energies. electric force, like the gravitational force fact, conservative. This means that the potential energy D is,–in Electric potential energy and p.d. difference does NOT depend on the path of the charge. 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 kq0Q Ue = (use the point charge form) r0 rf Ue = kq0Q 1 - 1 rf r0 1 1 Ue = (9×109)(-1.6×10-19)(1.6×10-19) 2 1 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? Topic 6.2 Extended D – 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: Ue V = q0 = +1.152×10-28 J -1.6×10-19 C = -7.2×10-10 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 - kQ Ue = V = r0 q0 rf Topic 6.2 Extended D – 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: UE = q0Edf - q0Ed0 UE = q0E(df - d0) (use the parallel plate form) UE = (-1.6×10-19)(250)(0.02 - 0) UE = -8×10-19 J Topic 6.2 Extended D – 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 mv2 2 1 (9.11×10-31)v2 2 = 8×10-19 = 8×10-19 v = 1.3×106 m/s FYI: Don’t confuse voltage V with velocity v. Topic 6.2 Extended D – 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: V = Ed (use the parallel plate form) V = (250)(0.02) V = 5 V FYI: The NEGATIVE sign means that these charges have a lower potential energy thanTopic they would6.2 if theyExtended were farther away. The highest potential for these particular charges energy would be ZERO, . D – energy Electric potential andat p.d. FYI: If all the charges were POSITIVE (or NEGATIVE), the total Find the total electrostatic potential energy would be positive. The lowest potential q3 =energy -2Cwould potential energy in ZERO. the occur at , and would be charge configuration shown. Assume the charges have been assembled from infinity! q1 = +5C q2 = +1C 0.25 m Pick the right formula and work in pairs: 0 kqaqb kqaqb Ue = (use the point charge form) rf kq1q2 r12 = k51 0.25 = 0.18 J kq1q3 r13 = -k52 0.25 = -0.36 J kq2q3 r23 = -k12 0.25 = -0.072 J U12 = U13 = U23 = U12 U13 U23 U12 U13 Utotal = 0.18 J + -0.36 J + -0.072 U23 J = -0.252 J Question: How would you find their speeds after they have separated Topic 6.2 Extended to infinity? Find their speed now. Is it what you expected? D – 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 kqq 1 1 mv2 + mv2 + = 2 2 rf r0 1 1 mv2 = kq2 r0 rf kq2 1 1 2 v = m r0 rf (9×109)(-1.6×10-19)2 1 1 = 9.11×10-31 .01 .10 v = 150.87 m/s v2