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90 101 N Spheres A and C have like charges, which repel The force of C on A is upward Find the vector sum of FB on A and FC on A to find Fnet on sphere A Fnet FB on A2 FC on A2 (90 101 N)2

Are the units correct (N m2/C2)(C)(C)/m2 N The units work out to be newtons Does the direction make sense Like charges repel; unlike charges attract Is the magnitude realistic The magnitude of the net force is in agreement with the magnitudes of the component forces



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9 A negative charge of 20 10 4 C and a positive charge of 80 10 4 C are separated by 030 m What is the force between the two charges 10 A negative charge of 60 10 6 C exerts an attractive force of 65 N on a second charge that is 0050 m away What is the magnitude of the second charge 11 The charge on B in Example Problem 1 is replaced by a charge of 300 C Diagram the new situation and find the net force on A 12 Sphere A is located at the origin and has a charge of 20 10 6 C Sphere B is located at 060 m on the x-axis and has a charge of 36 10 6 C Sphere C is located at +080 m on the x-axis and has a charge of 40 10 6 C Determine the net force on sphere A 13 Determine the net force on sphere B in the previous problem





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As you use the Coulomb s law equation, keep in mind that Coulomb s law is valid only for point charges or uniform spherical charge distributions That is, a charged sphere may be treated as if all the charge were located at its center if the charge is spread evenly across its entire surface or throughout its volume If a sphere is a conductor and another charge is brought near it, the charges on the sphere will be attracted or repelled, and the charge no longer will act as if it were at the sphere s center Therefore, it is important to consider how large and how far apart two charged spheres are before applying Coulomb s law The problems in this textbook assume that charged spheres are small enough and far enough apart to be considered point charges unless otherwise noted When shapes such as long wires or flat plates are considered, Coulomb s law must be modified to account for the nonpoint charge distributions

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There are many applications of electric forces on particles For example, these forces can collect soot in smokestacks, thereby reducing air pollution, as shown in Figure 20-12 Tiny paint droplets, charged by

492 ft 2139 MHz 23 ft 3 Director (K 4615): L K ft FMHz 4615 ft 2139 MHz 2158 ft 4 Element Spacing (K 142): L K ft FMHz 142 ft 2139 MHz 664 ft The elements of a rotatable beam antenna can be built in a manner similar to the rotatable dipole described earlier In the beam antenna, however, a boom is needed between the elements to support them The boom can be made of metal or wood In the case of a metal boom, the driven element must be insulated from the boom, even though the parasitic elements can be mounted directly to it In general, it is usually better to use wood as a matter of convenience Metal boom antennas can be obtained from commercial sources The wood boom is easy to build and maintain, even though a little less durable than a metal boom The feedpoint impedance of a dipole is on the order of 72 in free space, although the actual impedance will vary above and below that figure for antennas close to the earth s surface In addition, adding parasitic elements reduces the impedance even more The feedpoint impedance of the antenna is too low to be directly fed with coaxial cable, so some means of impedance matching is needed Some people feed the antenna through an impedance matching balun transformer Figure 12-4 shows the gamma match system The driven element of the Yagi is not

As shown in the figure on the right, two spheres of equal mass, m, and equal positive charge, q, are a distance, r, apart 1 Derive an expression for the charge, q, that must be on each

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