Chapter 19: Electric Charges, Forces, and Fields

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1 Ch. 0 Solution PHY 05 Hayel Shehadeh Chapter 9: Electric Charges, Forces, and Fields Answers Conceptual Questions 4. The two like charges, if released, will move away from one another to infinite separation, converting the positive electric potential energy into kinetic energy. The two unlike charges, however, attract one another if their separation is to be increased, a positive work must be done. In fact, the minimum amount of work that must be done to create an infinite separation between the charges is eual to the magnitude of the original negative electric potential energy. 6. The electric field is either zero on this surface, or it is nonzero and perpendicular to the surface. If the electric field had a nonzero component parallel to the surface, the electric potential would decrease as one moved in the direction of the parallel component. 0. Capacitors store electrical energy in the form of an electric field between their plates. This stored energy can remain in a capacitor even when a device such as a television is turned off. Accidentally touching the terminals can cause the sudden and potentially dangerous release of this energy. Solutions to Problems and Conceptual Exercises. A 4.5 µc charge moves in a uniform electric field E4.0 5 N/C The change in electric potential energy of a charge that moves against an electric field is given by euation 0-, U 0Ed. If the charge moves in the same direction as the field, the work done by the field is positive and U W Fd 0Ed. If the charge moves in a direction perpendicular to the field, the work and change in potential energy are both zero. The change in potential energy from point A to point B is independent of the path the charge takes between those two points. Solution:. (a) The charge moves perpendicular to the field:. (b) The charge moves in the same direction as the field: U 0 xˆ 6 5 U 0Ed 4.50 C 4.0 N/C 6.0m J.

2 Ch. 0 Solution PHY 05 Hayel Shehadeh. (c) The potential energy changes by an amount eual to the sum of paths and above: U 0 J J Note that U changes only along the direction parallel to the field. This is consistent with statement, The electric field points downhill on the potential surface, as described by figure 0- and the accompanying text. 8. An ion loses electric potential energy as it is accelerated through a potential difference. The electric potential is the electric potential energy per charge (euation 0-). To find the charge we need only divide the electric potential energy by the electric potential. Solution: Solve euation 0- for 0 : 5 U.7 0 J 0 V 40 V C 4.00 e If the charge were positive, its electric potential energy would increase as it moves from a lower to a higher potential. In this problem the negative charge loses electric potential energy and gains kinetic energy as it accelerates. A uniform electric field E 00 N/C xˆ creates a change in the electric potential along the ˆx direction. The change in electric potential is given by euation 0-4, V E s, where s is eual to x because the field points in the ˆx direction. There is no change in the electric potential along the yˆ and zdirections. ˆ The change in electric potential between any two points is independent of the path taken between those two points. Solution:. (a) x 0 between points A and B:. (b) Solve euation 0-4 for V between points B and C: V VB VA E x 0 V V V Ex E x x B C B C 00 N/C 0.040m 48 V. (c) Repeat for points C and A: V VC VA Ex E xc xa 00 N/C 0.040m 48 V 4. (d) No, we cannot determine the value of the electric potential at point A because the location of zero potential has not been established. We could always choose

3 Ch. 0 Solution PHY 05 Hayel Shehadeh VA 0 if we wish! Note that the potential changes only along the direction parallel to the field and is higher at point C than at either point A or point B. This is consistent with statement, The electric field points downhill on the potential surface, as described by figure 0- and the accompanying text. 6. The kinetic energy of an electron changes as it moves to locations of different electric potential. As the electron moves from place to place, its kinetic energy increases as it moves toward higher potential (remember, it is negatively charged, so it is accelerated opposite the field and toward positive charges) and decreases as it moves toward lower potential. Use the conservation of energy to determine the potential at point C and the kinetic energy at point A. Solution:. (a) Set Ki Ui Kf Uf for the first case:. Now set Ki Ui Kf Uf for the second case:. Solve the first euation for KA UA UC and substitute into the second, and solve for U : C 0 UA KA UC 0U K U B A C U U U U B A C C U U U C A B 4. Divide both sides by e to find V : ev C C eva evb V V V V 49 V 55 V C A B 5. (b) Substitute the result from part (a) KA UA UC eva VC into the expression from step to find K : e 55 V 8 ev A In the first case the electron moves from V (point A) to 55 V (point C), gaining 8 ev of kinetic energy, but in the second case the electron moves from 49 V (point B) to 55 V (point C), gaining 66 ev of kinetic energy.. The speed of a proton is changed as it moves through a potential difference. Use conservation of energy, with the potential energy of the proton given by U V ev, to determine the potential difference reuired to change the proton s speed and kinetic energy. Solution:. (a) Set Ki Ui Kf Uf and solve for V : mv ev mv ev i i f f mv mv e V V ev i f f i 7 5 i vf 9 e.600 C m v.670 kg 4.00 m/s 0 V 840 V 0.84 kv

4 Ch. 0 Solution PHY 05 Hayel Shehadeh. (b) Find V for when v v : f. (c) Set K U K U and solve for V : i i i f i m mv mv V v v e e 4 8e i i i i C.67 0 kg m/s 60 V 0.6 kv K ev K ev i i i f K i ev kg4.00 m/s 9 Ki mvi V e 4e C 0.4 kv It reuires a smaller potential difference to cut the kinetic energy in half than it does to cut the speed in half because the kinetic energy is proportional to the speed suared; cutting the speed in half reduces K by a factor of four. 0. Four different arrangements of point charges are shown at right. In each case the charges are the same distance from the origin. The electric potential at the origin is the sum of the potentials from each individual charge. The electric potential at the origin is proportional to the sum of the charges, because each charge is the same distance from the origin. Use this observation to determine the ranking of the electric potentials at the origin. Solution: The sum of the charges in the four cases are: A, 0; B, 0; C, ; and D, +. Therefore, the ranking of electric potential at the origin is as follows: C < A = B < D. The electric potential a long distance from the origin also follows the same ranking, because from a long distance the charges appear as a single charge with a magnitude eual to the sum of the charges. 4

5 Ch. 0 Solution PHY 05 Hayel Shehadeh 7. Three charges are arranged at the corners of a rectangle as indicated in the diagram at right. The work reuired to move the.7 µc charge to infinity is eual to the change in its electric potential energy due to the other two charges. Sum the electric potential energies due to the other two charges to find the electric potential energy at the corner of the rectangle. At infinity the electric potential energies are zero. Use these facts to find the work reuired to move the.7 µc charge to infinity, and then repeat the procedure to find the work reuired to move the 6. µc charge. Solution:. (a) Find U U U 0 U U : f i k k W U 0 k r r r r N m C.0 C C C 0.5m (0.5m) (0.6 m) W 0.86 J. (b) The 6. C charge is repelled by the. C charge more than it is attracted by the.7 C charge. The work reuired will be negative, which is less than the work reuired in part (a).. (c) Repeat step for the 6. µc charge: k k W U 0 k r r r r N m C 6.0 C 0.5m 0.6 m W 0.54 J C.0 C Note that the work done by the electric field is W U (euation 0-) but the work you must do to move the charge against the field is W U. The negative work of part (c) means the charge would fly off on its own. 47. The figure at right shows a series of euipotentials in a particular region of space, and five different paths along which an electron is moved. Euipotential lines always intersect electric field lines at right angles. Use this principle together with the concepts of electric field intensity and work done by an electric field to answer the uestions. Solution:. (a) The value of the electric potential decreases as we move in the direction of the electric field (see, for example, figure 5

6 Ch. 0 Solution PHY 05 Hayel Shehadeh 0-.) This means that the electric field must point to the left, corresponding to a decrease in the electric potential from 60 V to 5 V.. (b) Recall that the force exerted on an electron is opposite in direction to the direction of the electric field; therefore, the electric field does positive work on an electron that moves opposite in direction to the electric field, because the force and the direction of motion are the same. This means that any path in the diagram that moves the electron toward the right corresponds to positive work done on the electron by the electric field. It follows that the works done along the paths in the figure have the following signs: A, positive; B, positive; C, positive; D, negative; E, negative.. (c) The work done by the electric field is proportional to the difference in electric potential between the starting and ending points of the path. In fact, euation 0- says that W 0V ev e V for these electrons. The V for each case is A, +5 V; B, +40 V; C, +5 V; D, 5 V; E, 5 V. The ranking is thus: E < D < A < C < B. 4. (d) The magnitude of the electric field is proportional to the magnitude of the potential difference divided by the distance. The distance for paths A and E is the same, but the magnitude of the potential difference is much greater for path E than for path A. It follows that the electric field near path A is less than the electric field near path E. The work done by the electric field is independent of the particular path because the electric force is a conservative force. The work depends only upon the endpoints of the path. 58. A capacitor is formed with two parallel metal plates separated by a small distance. Euation 0- gives the relationship between the capacitance of a parallel-plate capacitor, the plate area A, and the plate separation d. Substitute this expression into euation 0-9, C Q V, and solve for Q. The air between the plates has a dielectric constant (.00059) that is close enough to one that we can approximate it as a vacuum. Solution:. (a) Solve euation 0-9 for Q and substitute euation 0- for C: Q AV C m V 0 Nm CV d m.6 nc. (b) Because Q is inversely proportional to d, the answer to part (a) will decrease if d is increased.. (c) Repeat part (a) with the new d: C ( m )( V) Nm 0 Q 7.80 C m 6

7 Ch. 0 Solution PHY 05 Hayel Shehadeh The battery maintains a constant potential difference V between the plates, so that E = V / d must decrease if d is increased. In order to do that, the charge density and therefore charge Q must decrease ( E 0, see active example 9-). 8. Three charges are arranged as shown at right. The electric potential at the origin is the sum of the potentials from each of the three charges. Sum the three potentials and set the sum eual to zero, then solve the resulting expression for. Solution:. Sum the potentials:. Solve for : V 0 k r r r r r r. Find r, r, and r : r 4.40 m 6. m 7.6 m, r 4.50 m 6.75 m ( 4.50 m, 6.75 m) r. m. m.99 m y x O (. m,. m) 4 6 (4.40 m, 6. m). Insert the numerical values: 4.5 C. C.99 m 7. C r r r 7.6 m 8. m A third charge of relatively small magnitude is reuired because that charge is closer to the origin than the other two charges. 7