(Solution Manual) College Physics Reasoning and Relationships (Volume 2)

Giordano, 2

Physics

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2nd Edition

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Giordano: CP 2e ISM, Vol 2 1 CH17 17 Electric Forces and Fields CONCEPT CHECKS 17.1 | Superposition and the Direction of the Electric Force The correct answer is (b). Electron 1 (on the left) repels electron 2, while the proton at- tracts electron 2. Since all the charges have the same magn itude and the proton is closest, its force is largest in magnitude, so the net force on electron 2 is to the left, along the −x direction. 17.2 | Force on a Point Charge The correct answer is (a). Suppose Q is negative. If placed in region A, it will be at tract- ed to q and repelled by the charge −2q. Since q is closer, there will be a point in r egion A where these two forces are equal in magnitude and thus cancel. A similar argument applies if Q is positive. 17.3 | Holding Together Two Charges The correct an swer is (a). The force between two charges with q = +1 C separated by a distance r = 1 m has a magnitude 17.4 | Electric Forces and Neutral Objects Yes—it is possible. Two such possibilities are these arrangements of electric di poles: 17.5 | Direction of an Electric Field The correct answer is (d). At the origin, the electric fields from the charges on the x axis will be in opposite directions and cancel. The field from the positive charge on the y axis will be in the –y direct ion. Giordano: CP 2e ISM, Vol 2 2 CH17 17.6 | Force on a Polarized Object The correct answer is (b). Changing the charge on the rod from negative to positive will change the direction of the electric field and also reverse the direction of polarization, so the side of the water nearest the rod will now be negatively charged. The water will thus still be attracted to the rod. 17.7 | Finding the Electric Flux The correct answer is (c). The total electric flux is zero because the total charge en- closed by this surface is zero. In Figure 17.30, each field line that enters the surface al- so exits the surface. 17.8 | Electric Field from a Line of Charge The correct answer is (b). Symmetry requires that be perpendicular to the line (for the same reasons given in the discus sion of parts B through E of Figure 17.31). Be- cause is directed toward a negative charge, the field is toward the line. QUESTIONS Q17.1 If we rub the glass rod with the silk, the glass rod will become positively charged. If the charged object is attracted to the glass rod, then the object is charged negative (unlike charges attract). If the charged object is repelled by the glass rod, the n the ob- ject is charged positive (like charges repel). Similar tests could be made with the silk. Q17.2 The electric field vector at any point shows the direction of force on a positive charge at that point. If electric field lines crossed, it would mean a charge placed at that point could be pushed in either one of two different directions! Q17.3 Electric field lines start at positive charge and terminate on negative charge, so charges a and c are positive, and charge b is negative. Q17.4 The density of f ield lines is proportional to the magnitude of the field. By Cou- lomb’s law, larger charges create proportionally larger electric fields. This means that larger charges have a higher density of field lines near them. By counting the number of field lines th at start or terminate on each charge, we determine that the largest charge is charge b, and the smallest charge is charge c. [SSM] Q17.5 The uncharged object has molecules which contain an equal number of positive and negative charges. When the uncharged object is brought near the charged object, the neutral molecules of the uncharged object will become electrically polarized. Giordano: CP 2e ISM, Vol 2 3 CH17 For example, when a positively charged object is brought near a zero -net-charge object, the electrons in the zero -net-charge object (which are more mobile than the positive charges) move slightly toward the positively charged object. Since the electrons have moved clo ser, the zero -net-charge object is slightly negative on the side closest to the positive object (and slightly positive on the side away from the positive object). This “in- duced” charge causes an attraction between the objects. Q17.6 Zero net charge can still mean both objects to have one end that is slightly posi- tive and other end slightly negative; the objects could be electrically polarized. Consider two possible ways this could happen: (a) The neutral (no net charge) molecules in the two objects could be subjected to an external electric field. When this happens, the electrons in the neutral molecules (which are more mobile than the positive charges) move slightly in the direction op- posite the direction of the external electric fie ld. Now that we have a slight separation of charge in the molecules of both neutral objects, they can attract each other. (b) The objects might both be polarized insulators, which means that each end could be charged. For example, we could have two rod -shaped insulators, each with an ex- cess of charge at one end but each with zero net charge. If the positive end of one insulator is placed near the negative end of the other, the two objects will attract each other. Q17.7 The glass rod that has been rubbed wit h silk is positively charged. Since the charged point particle is repelled by the glass rod, it must be positive (like charges re- pel). Q17.8 The rubber -like material of the balloon acquires a negative charge, and it is also an insulator. When a balloon is placed near a neutral wall, the outer layer of the wall material becomes polarized by induction. (Electrons in the wall move away slightly, cre- ating a small positive charge area on the surface.) The balloon may lose some charge at the point of contact (ele ctrons lost to the wall), but since the balloon material is a very poor conductor, the rest of its charge remains on the balloon and induces enough polar- ization to create enough electric force to hold the balloon in place. The electric attra ction produces a normal force on the balloon, pushing it against the wall so that upward fric- tion then holds the balloon against gravity. Q17.9 The charged rod has an excess of electrons. When the rod is brought close to the pellets the neutral pellets become polarized. The positive end of each pellet is pulled (accelerated) toward the rod by the Coulomb interaction. When the pellet makes contact with the rod, some electrons move on to the pellet. The pellet now has a net negative Giordano: CP 2e ISM, Vol 2 4 CH17 charge, so it is repelled by and acceler ated away from the rod, again by the Coulomb interaction. Q17.10 (a) The two charges Q have the same charge and are equidistant from the charge q. As a result the two Coulomb forces on q are equal in magnitude and opposite in direction. Adding these together, the total force on q is zero. (b) If q has the same charge a s Q and it is moved away from the origin, it is closer to one of the charges than the other. The repulsion force is therefore greater than the a t- tractive force, and it is pushed back toward the origin. If q has the opposite charge as Q and it is moved away from the origin, it will continue in the direction it was moved. For instance, if the charge is moved in the + x direction, the attractive force from the charge along the positive x axis grows larger than the attractive force from the charge along the –x axis, which keeps the charge moving in the + x direction. (c) If q is in stable equilibrium, you can displace it slightly and it will oscillate about the equilibrium position. The charge q is in stable equilibrium when it has the same charge as Q, but in uns table equilibrium when it has a charge opposite that of Q. Q17.11 Since electric fields terminate on negative charges, the direction of the electric field near the surface of the Earth will be toward Earth’s surface. Q17.12 Electrons are external to the nucleus. Outer electrons are not only a considera- ble distance (rela tively speaking) from the nucleus but they also receive a considerable amount of shielding from the nucleus by the inner electrons, so it is easy for them to move about. By contrast, the protons are tightly bound within the heavy nucleus and are not free t o move about. Q17.13 The total mass of the dust particle is: If we assume that the dust particle is carbon which has a molecular mass of: The number of moles of carbon dust is: Giordano: CP 2e ISM, Vol 2 5 CH17 The number of carbon dust atoms is: This estimate is within one order of magnitude of the given number of electrons. Q17.14 The children are touching the Van De Graff generator which has a positively charged dome. Since the dome is positively charged, electrons leave the children and move onto the generator dome. This leaves the children with a net positive charge (hair and all). Since like charges repel, the children’s hair stands out as one strand repels another. [SSM] Q17.15 Three main simi larities come from the form of the force equations: Both have a proportionality constant ( k and G), both are proportional to a property of the in- teracting bodies (mass and charge), and most importantly they both have magnitudes that decrease with the inver se square of the distance between the interacting objects. The relative strengths for these forces are, however, very different. This can be seen by comparing the sizes of G and k. Also, gravity is always attractive, for any kind of mass. This stands in s harp contrast to the Coulomb force where two kinds of charge can r esult in either attractive or repulsive forces. Q17.16 If the test charge is not small then the density of field lines from the test charge would change the fields from the source in the re gion around the test charge. This al- tered field would be different than the field of the source charge alone. Q17.17 The flux does not change. Flux is a measure of the number of field lines pene- trating a surface area (actually the perpendicular component o f these field lines). The charge in the balloon has a certain number of field lines associated with it. For example, consider a positive charge with 100 field lines inside of a spherical balloon. These 100 field lines all originate on the positive charge a nd come through the surface of the ba l- loon. If the balloon is small the 100 field lines are very close together over the small ar- ea. If the balloon goes to a larger volume and surface area, the same 100 lines still come through, but there are fewer lines p er unit area. No matter what size the balloon, the number of field lines coming though, which indicates the flux, is the same. Q17.18 Answers will vary. The main idea here is to consider that charge doesn’t have to come in a minimum quantity for Coulomb’s and Gauss’s laws to apply, but that with-Giordano: CP 2e ISM, Vol 2 6 CH17 out quantization, it is very difficult to conceptualize charge. For instance, an atom or a normal piece of an element is usually electrically neutral. How could an atom be pre- cisely neutral if charge is not quantize d? When a spark jumps across an air gap, what moves? A continuous stream of stuff? Is there any mass associated with the movement of charge, and if so, how can mass be continuous? This example shows just one way that conceptualization of a concept (like ch arge) dramatically affects our ability to pr edict electrical effects. Q17.19 The direction of the electric field shows the direction of force on a positive charge. If the ion moves in a direction opposite the field, the force must also be opposite the fiel d, which would indicate a negatively charged particle. Q17.20 Field lines originate at positive charge and terminate at negative charge. In case 3, all of the field lines are leaving the cube, and therefore they all must originate within the cube. This imp lies the cube in case 3 contains a positive charge. In case 2, six field lines enter the box (implying negative charge), but only two field lines leave (implying positive charge). This implies the net charge in case 2 is negative. In case 1, an equal numbe r of field lines enter and leave, implying an equal amount of positive and negative charge, or zero net charge. PROBLEMS P17.1 Recognize the principle. The total charge can be found from the charge on one electron and the number of electrons in one mole. Sketch the problem. No sketch needed. Identify the relationships. Knowing the charge on a single electron ( e) and that there are Avogadro’s number ( NA) of electrons in a mole of electrons, it is possible to deter- mine the charge of one mole of electrons by: Q = eNA Solve. The charge on Avogadro’s number (one mole) of electrons is therefore: Q = eNA = (−1.6 × 10−19 C)(6.02 × 1023) = −9.6 × 104 C What does it mean? Since typical static electrical charges we experience on a daily ba- sis are on the order of microcoulombs, this is a very large charge by comparison! P17.2 Recognize the principle. We can find the number of moles of hydrogen (n) from the total mass ( mtotal ) and the molecular mass ( Mhydrogen ), and then determine the num-Giordano: CP 2e ISM, Vol 2 7 CH17 ber of molecules ( N) using Avogadro’s number ( NA). Finally, we can determine the number of protons and electrons from the number of molecules, and the total ch arge from each of these quantities. Sketch the problem. No sketch needed. Identify the relationships. From the periodic table, 1 mole of hydrogen has a mass of: Mhydrogen = 1.0 g/mole = 0.0010 kg/mole Each molecule of hydrogen consists of two atoms of hydrogen and hence two electrons and two protons. The charg e on each electron is –1.60 × 10–19 C, while the charge on each proton is +1.60 × 10–19 C. The number of moles of hydrogen may be determined by: The number of molecules of hydrogen may be determined by: N = nNA Solve. Combining these two expressions and inserting values gives: Since each hydrogen atom has one electron and one proton, the number of electrons and/or protons in the 5 g of hydrogen is: Nelectrons = Nprotons = 3.00 × 1024 protons/electrons The total charge on all the electrons in 5 g of hydrogen is: Qelectrons = (–e)(Nelectrons ) = (–1.60 × 10–19 C)(3.00 × 1024) = –4.80 × 105 C In a similar way, the total charge on all the protons in 5 g of hydrogen is: Qprotons = +4.80 × 105 C What does it m ean? This is a very large amount of charge, but is generally neutralized Giordano: CP 2e ISM, Vol 2 8 CH17 because both positive and negative charges are present. The “charge” we usually see on a material is due to addition or removal of a comparatively small part of this total charge. P17.3 Recognize the principle. The total charge indicates the number of electrons. Both the amber and the fur start electrically neutral, and the amber gains a negative charge equal and opposite to the positive charge on the fur as the two are rubbed to- gethe r. Sketch the problem. No sketch needed. Identify the relationships. Per Equation 17.1, the charge on one electron is: e– = –1.60 × 1019 C Solve. We can therefore multiply the given charge by the ratio of the magnitude of the charge to find the number of e lectrons: What does it mean? Creating a charge even on the order of nanocoulombs still in- volves moving a large number of electrons. [SSM][Reasoning] * P17.4 Recognize the principle. We can make this estimate from our total weig ht (and therefore mass) by assuming that our body is mostly water. Sketch the problem. No sketch needed. Identify the relationships. The total mass ( mtotal) of a person is about 70 kg. From the periodic table, the molecular mass ( Mwater) of water is 18 g/m ole. The number of moles of water (n) may be determined by: The number of molecules ( N) of water may be determined by: N = nNA There are 10 electrons (8 from oxygen and 1 in each hydrogen) in each water molecule, and the charge on each electron is –1.6 × 10–19 C. Giordano: CP 2e ISM, Vol 2 9 CH17 Solve. Combining the mole and molecules expression, the number of molecules of wa- ter in our body is: With 10 electrons in each water molecule, the total number of electrons is: Nelectrons = 10N = 10(2.34 × 1027) = 2.34 × 1028 The tota l charge on these electrons is: Qelectrons = (–e)(Nelectrons ) = (–1.60 × 10–19)(2.34 × 1028)= –3.74 × 109 C What does it mean? Again, this is a very large charge that is not generally apparent because an equal amount of positive charge is also present. P17.5 Recognize the principle. The net charge will be determined by the excess (or unmatched) electrons. Sketch the problem. No sketch needed. Identify the relationships. We know there is an excess of 45 electrons and, from Equa- tion 17.1, we know the charge of each electron is −1.6 × 10–19 C . We can therefore de- termine the net charge by multiplying the charge on an electron by the number of ex- cess electrons. Solve. The charge can then be written as: Q = –eN = (–1.60 × 10–19 C)(45) = −7.2 × 10–18 C What does it mean? The charge on an object is determined by the number of un- matched electrons. [Reasoning] * P17.6 Recognize the principle. We can estimate the mass of a penny. From this mass, we can find the number of moles and then molecules of copper. The number of electrons can then be found from the number of molecules. Sketch the problem. No sketch needed. Identify the relationships. The mass of a penny is mtotal = 1.00 g = 0.001 kg. From the Giordano: CP 2e ISM, Vol 2 10 CH17 periodic table, the molecular mass of copper is Mcopper = 63.5 g/mole = 0.06035 kg/mole. The number of moles of copper ( n) may be determined by: The number of atoms ( N) of copper may determined from Avogadro’s number ( NA): N = nNA From the periodic table, each copper atom contains 29 electrons. Solve. Combining the number of moles and Avogadro’s number, the number of a toms is: Since each copper atom contains 29 electrons, the total number of electrons is: Nelectrons = 29N = 29(9.48 × 1021) = 2.75 × 1023 electrons What does i t mean? The charge of this large number of electrons is balanced by an approximately equal number of protons in the penny. P17.7 Recognize the principle. We can find the total charge on the given area from the charge density, and then determine the number of electrons from the charge. Sketch the problem. No sketch needed. Identify the relationships. The charge density  for a charge q spread over an area A is defined as: We know that 1 mm2 = 1 × 10−6 m2, and Equation 17.1 tells us that each electron has a charge of –1.60 × 10–19 C. Solve. Solving this relationship for the charge and inserting values, Giordano: CP 2e ISM, Vol 2 11 CH17 q =  A = –0.005 C/m2(1.0 × 10–6 m2) = –5.0 × 10−9 C Then, the number of electrons is: What does it mea n? A large number of electrons must be added in even a small area to create this charge density. P17.8 Recognize the principle. The number of excess electrons can be determined from the total charge. Sketch the problem. No sketch needed. Identify the relat ionships. The total charge on the rod is simply the charge per unit length times the length, or: q = L Each electron has a charge of −1.6 × 10−19 C. Solve. Inserting values, the total charge is then: q = (−7.5 × 10−6 C/m)(2.5 m) = −1.86 × 10−5 C This impl ies a total number of electrons of: What does it mean? Even a charge in the tens of microcoluombs involves the dis- placement of hundreds of trillions of electrons! P17.9 Recognize the principle. A positively charged ion comes fr om a loss of elec- trons. Sketch the problem. No sketch needed. Identify the relationships. A calcium ion with a +2 charge has two more protons than electrons. Each extra proton has a charge of 1.6 × 10−19 C. Solve. The total charge in coulombs is therefore: Giordano: CP 2e ISM, Vol 2 12 CH17 2(1.6 × 10−19 C) = 3.2 × 10−19 C What does it mean? The charge on an individual ion is very small when measured in coulombs. P17.10 Recognize the principle. The magnitude of the force between two charged particles is determined by Coulomb’s law. Sketch t he problem. No sketch needed. Identify the relationships. Coulomb’s law, for two charged particles with charges q1 and q2 and separated by a distance r is given in Equation 17.3: where . The magnitude of the charge on each electron is 1.60 × 10−19 C. Solve. Inserting the given values, What does it mean? This force seems very small, but is actually rather large on an atomic scale. If this were the only force acting on the ele ctron, it would create an accel- eration of 2.5 × 1022 m/s2! P17.11 Recognize the principle. The magnitude of the electric force on charged parti- cles separated by a distance is determined by Coulomb’s law. Sketch the problem. No sketch needed. Identify the relationships. Coulomb’s law, for two charged particles with charges q1 and q2 and separated by a distance r is given in Equation 17.3: where . Giordano: CP 2e ISM, Vol 2 13 CH17 A proton has a charge q1 = +1.60 × 10–19 C, and an electr on has charge q2 = –1.60 × 10–19 C. Solve. The magnitude of the force F is therefore given by: What does it mean? Because they have opposite charges, the force between the elec- tron and proton is attractive. On an atomic scale, a force with this magnitude is very sig- nificant! P17.12 Recognize the principle. The force on charged particles separated by a dis- tance is determined by Coulomb’s law. Sketch the problem. No sketch needed. Identify the relationships. Coulomb’s law gives the force F on two particles separated by r as: (a) Solve. Inserting the given values: (b) Solve. If the charge values are negative, this becomes: What does it mean? The answe r does not change if we switch the sign of both parti- cles. The negative sign in the above answer tells us that the force between the two charged particles is attractive. * P17.13 Recognize the principle. The electric force on two charged particles sepa- rated by a distance is determined by Coulomb’s law. Sketch the problem. See Figure P17.13. Giordano: CP 2e ISM, Vol 2 14 CH17 dentify the relationships. Coulomb’s law states that the magnitude of the force is in- versely proportional to r2. This means as the distance ( r) gets greater the magnit ude of the force ( F) gets smaller. In fact if the distance doubles, the magnitude of the force should be one -fourth as large. Solve. Only plot (c) shows a force magnitude that gets smaller with the square of the distance. What does it mean? The electros tatic force between charged particles, like the gravita- tional force, follows an inverse square law. * P17.14 Recognize the principle. The magnitude of the force between two charged particles separated by a distance is given by Coulomb’s law. Sketch the pro blem. See Figure P17.14. Identify the relationships. Coulomb’s law gives the magnitude of the force F on two par- ticles separated by r as: The distance between the particles can be found using the given coordinates: Using Newton’s third law, the force of q1 on q2 is equal in magnitude and opposite in di- rection to the force of q2 and q1. Solve. The magnitude of the force between q1 and q2 is then: Since the charges have the sa me sign, this force is repulsive, and acts along a line be- tween the particles. This line has an angle (  ) with respect to the x axis such that: The force of q1 on q2 is 4.5 × 109 N in a direction 26.6° above the +x-axis . Giordano: CP 2e ISM, Vol 2 15 CH17 The force of q2 on q1 is equal in magnitude, but opposite in d irection to part (a), or 4.5 × 109 N in a direction 26.6° below the −x axis. What does it mean? Two like charges always repel each other according to Coulomb’s Law. [Reasoning] * P17.15 Recognize the principle. We can reason what the charges might be by considering symmetry and the typical direction of field lines. Sketch the problem. See Figure P17.15. Identify the relationships. By symmetry, the two charges must be the same, since a dif- ferent charge would result in a net upward or downward field. Also, since field lines point away from positive charges and toward negativ e charges, the two charges on the right should be negative, or, if they are positive, smaller than + Q so that the net field is to the right. Solve. The two charges should be equal and negative, or equal and positive, but smaller than + Q. What does it me an? It is important to develop a general understanding of how charges create fields. * P17.16 Recognize the principle. If the net force on q is to be zero, the charges Q and 5 Q must exert forces of equal magnitude and opposite direction on q. We can use Coulomb’s law to write expressions for the force on q due to Q and 5 Q in terms of k, q, Q, L, and . Sketch the problem. See Figure P17.16. Identify the relationships. The third particle (the one placed at the origin) has an un- known charge q. Expressions for the magnitude of the force on q due to Q (FqQ) and 5 Q (Fq5Q) are written using Coulomb’s law as: Solve. We set these forces equal to each other, and then solve for : Giordano: CP 2e ISM, Vol 2 16 CH17 What does it mean? We might have anticipated this; the ratio of the charges (5:1) is equal to the inverse ratio of the distances squared . * P17.17 Recognize the principle. The force on charged particles separated by a dis- tance is determined by Coulomb’s law. Sketch the problem. No sketch needed. Identify the relationships. When the particles are a distance r apart, the force F is de- termined by Coulomb’s law: We can write Coulomb’s law for both distances ( r and r’ = r/1.5), and then take the r atio of F’ and F. Solve. When the particles are a distance r’ apart (where r’ = r/1.5), the force F’ is de- termined by: And since: this implies that: What does i t mean? Decreasing the distance between two charged particles by a factor of 1.5 means the force between them is increased by a factor of 2.25. P17.18 Recognize the principle. The net force is a superposition of the Coulomb force Giordano: CP 2e ISM, Vol 2 17 CH17 from each charge. Sketch the problem. Figure P17.18 provides a good sketch. Identify the relationships. We must choose charges in each case that balance the hori- zontal and vertical components of force. (a) For point A, choosing the same sign for Q2 and Q3 results in horizontal s ymmetry since both will repel or attract the test charge. Q1 must then have an opposite charge, to result in an opposite vertical force that balances the force of both Q2 and Q3. (b) For point B, all charges must have the same sign in order to provide a b alanced at- traction or repulsion along both the horizontal and vertical axes. (c) For point C, Q2, and Q3 must have the same sign as they must provide balanced horizontal attraction or repulsion. Q1 must have the opposite sign in order to oppose the attraction or repulsion from Q2 and Q3. Solve. (a) iv (b) Either i or ii (c) iv What does it mean? The geometry allows for only certain charge combinations. Using the different distances to each point we could also predict relative magnitudes for each of these charges. P17.19 Recognize the principle. The force between charged particles is determined by Coulomb’s law. Sketch the problem. No sketch needed. Identify the relationships. Coulomb’s law, which tells the magnitude of the force be- tween two charged par ticles with charges q1 and q2 and separated by a distance r is given in Equation 17.3: where Giordano: CP 2e ISM, Vol 2 18 CH17 Solve. We can solve this equation for the distance between the particles, and insert values: What does it mean? The particles are about 3.4 cm apart. * P17.20 Recognize the principle. When the particles touch, the charge spreads equally over both conducting objects. We can find the new equal charges and then use Coloumb’s law to de termine the magnitude of the electric force between them. Sketch the problem. No sketch needed. Identify the relationships. When the two particles touch, the total charge is spread equally between the two of them. This total charge is: This implies that each particle has a charge of –1.0 µC after contact is made. We can then find magnitude of the force between two charged particles with charges q1 and q2 and separated by a distance r using Coulomb’s law, Equation 17.3: where Solve. Inserting our charge values and the given distance between the particles: What does it mean? The force between these particles is significant —about half the force of gravity on a typical adult! * P17.21 Recognize the principle. We can find the net force on each charge by adding the forces from the other two charged particles as vectors. Sketch the problem. See Figure P17.21. Identify the relationships. Since the horizontal distances and charges of the two parti- cles on the x axis are equal, by symmetry we can determine the forces along the x axis Giordano: CP 2e ISM, Vol 2 19 CH17 are equal and opposite. Since like charges repel, the net force is in a downward dire c- tion and equal to the sum of the two equal y-components of force. We can find the magnitude of the force using Coulomb’s law, as given in Equation 17.3: where The separation between the bottom particle and either particle on the x axis is then: And since the sides are equal, the downward component of force ( Fy) is related to the total force ( F) as Fy = F sin(45 °). Solve. Combining these expressions, our expression for the force acting on the bo ttom particle from one particle is: And, since bo th particles provide the same force, the total force is then: Inserting values, (b) Identify the relationships. The charge at the right is subject to a force from both the bottom charge and the left c harge. Since like charges repel, the force from the bottom charge is 45° above the horizontal, and the force from the left charge is directly to the right. The magnitude of each charge can be written as: and (b) Solve. These forces must be added as vectors, so summing force components in Giordano: CP 2e ISM, Vol 2 20 CH17 the x and y directions yields: Inserting values, Combining these gives a magnitude and direction of: What does it mean? The two side particles are subject to a slightly smaller net force magnitude than the bottom particle. If released, the bottom particle would accelerate downward, and the two side particles upward and away from each other. [SSM] * P17 .22 Recognize the principle. The net force on an electron at the origin would be the superposition of the forces from each of the three charges in proximity. Sketch the problem. See Figure P17.21. Identify the relationships. We can save ourselves considera ble work by realizing that since the two charges on the x axis have equal charge and are the same distance from the electron, each would attract the electron with equal and opposite force. There would therefore be no net horizontal attraction. The net forc e is therefore just a result of the charge on the –y axis. The magnitude of the charge can be determined by Coulomb’s law, given in Equation 17.3: where Solve. Inserting the charge and length values, the magnitude of the force is: Giordano: CP 2e ISM, Vol 2 21 CH17 The charge on the –y axis is positive, and so would attract the electron, resulting in a force in the –y direction. What does it mean? We can save ourselve s significant effort and calculation by realiz- ing the symmetry of this configuration of charges. [Reasoning] P17.23 Recognize the principle. We can use Newton’s law of gravitation to determine the magnitude of the gravitational force between the boxes fro m the mass of the electrons and the distance between them. Sketch the problem. No sketch needed. Identify the relationships. Since we know the total mass of each group of electrons and the distance between them, the magnitude of the force can be written as : Solve. Inserting values, the magnitude of the gravitational force between the two boxes of electrons is: We can compare the magnitudes of the electric and gravitational forces by taking their ratio: Finally, we can write: What does it mean? The gravitational force is extremely small (or weak) compared to the electric force. * P17.24 Recognize the principle. We can eliminate many of the attractiv e and repul-Giordano: CP 2e ISM, Vol 2 22 CH17 sive forces in this problem by symmetry, and then use Coulomb’s law to find the ma gni- tude of those that do not cancel. Sketch the problem. We construct a sketch showing all four forces acting on one of the dipoles. Figure Ans 17.24 Ident ify the relationships. F++ and F−− will have the same magnitude and will both be in the −x direction, the y components of F+− and F−+ will cancel, and the x components of F+− and F−+ will have the same magnitude and both be in the +x direction. We can find the sum of all forces in the x direction as given by Coulomb’s law to determine the elec- tric force between the dipoles. Solve. The distance between the like charges of the dipoles is given as 7.5 m. The magnitude of F+ + and F– – is, therefore: The distance between the unlike charges of the dipoles is: The magnitude of F+ – and F– + is, therefore: The angle  from the geometry in Figure P17.22 is: Giordano: CP 2e ISM, Vol 2 23 CH17 This implies the x component of F+ – and F– + is: Finally adding all x forces to find the net electric force between the dipoles, we have: What does it mean? The negative sign indicates that the net force is in the –x direction, away from the other dipole. Since the like -charged ends of the dipole are closer, a net repulsion is what we would expect. * P17.25 Recognize the principle. We can use Coulomb’s law to find the point at which the forces between each stationary charge and the newly placed charge are equal in magnitude and opposite in direction. Sketch the problem. Figure Ans 17.25 Identify the relationships. Since the sign on the charge Q and 3 Q is the same, q can be in equilibrium only if it is located somewhere between Q and 3 Q. If the total electric force on q is to be zero, then q must be closer to the smaller charge ( Q). We can define (as pictured) that q is some displacement d from the origin, an d then write the Coulomb force on q due to both Q and 3 Q. Finally, we can equate these forces and solve for the displacement d. Solve. The Coulomb force on q due to Q and 3 Q, respectively, are: Equating these forces we have: Giordano: CP 2e ISM, Vol 2 24 CH17 We can cancel k, q, and Q and simplify to obtain: (L + d)2 = 3(L – d)2 After performing the indicated algebra operations we obtain a quadratic in d: d2 – 4Ld + L2 = 0 This quadratic may be solved to obtain: If we use the positive root we obtain a number larger than L (3.73 L). This is contrary to our assertion that q must be between th e two charges. If we use the negative root we obtain a number smaller than L (d = 0.27 L) and this is in agreement with our thinking that q must be between the two charges. Finally since we know that q must be closer to Q, d must be a negative quantity, so we obtain: What does it mean? In this case, we need to apply both mathematics and our intuition about charges in order to arrive at the correct answer. The other mathematical root makes physical sense only if q carries a charge opposite to Q and 3 Q. * P17.26 Recognize the principle. If we choose a point anywhere along a line be- tween the two points, the electric forces from the two point charges will be in opposite directions. We can then use Coulomb’s law to match the magnitudes of the forces on the electron or proton along this line. Sketch the problem. No sketch needed. (a) Identify the relationships. The total distance between the two placed charges is: If we place the electron or proton along a li ne between the two points at a distance d from the origin, then the distance to the other point is 2.8 − d. We can then write an ex- pression balancing the two electric forces. The charges of the signs are dropped be- cause we are considering only magnitudes, and using Coulomb’s law, the force expres-Giordano: CP 2e ISM, Vol 2 25 CH17 sion is: (1) (a) Solve. Solving this expression for d yields: (2) Using the quadratic formula, the only positive root of this expression is d = 0.99 m. (The negative root would give a second point along the same line, but not between the particles.) The line between the particles is at an angle above the x axis of: This impli es the x and y coordinates of the point are: (b) Identify the relationships. The charge of the electron cancelled in the equation la- beled (1) above, which implies that the charge of the inserted particle makes no differ- ence. (b) Solve. The equilibrium point for a proton would be the same as that for an ele ctron, that is, (x = 0.45 m, y = 0.93 m) (c) Identify the relationships. The ratio of the charges of the two point particles was 3.33 originally, as indicated in Equation (2) above. Doubling both charges results in the same ratio. (c) Solve. If both charges are doubled, the equil ibrium point for a proton or electron would remain the same, that is, (x = 0.45 m, y = 0.93 m)