Solutions for HW chapter 18
2- REASONING The law of conservation of electric charges states that the net electric charge of an isolated system remains constant. Initially the plate-rod system has a net charge of 3.0μC + 2.0 μC = 1.0 μC. After the transfer this charge is shared equally by both objects, so that each carries a charge of 0.50 μC. Therefore, 2.5 μC of negative charge must be transferred from the plate to the rod. To determine how many electrons this is, we will divide this charge magnitude by the magnitude of the charge on a single electron.
SOLUTION The magnitude of the charge on an electron is e, so that the number N of electrons transferred is
4- REASONING The law of conservation of electric charges states that the net electric charge of an isolated system remains constant. Initially the plate-rod system has a net charge of 3.0μC + 2.0 μC = 1.0 μC. After the transfer this charge is shared equally by both objects, so that each carries a charge of 0.50 μC. Therefore, 2.5 μC of negative charge must be transferred from the plate to the rod. To determine how many electrons this is, we will divide this charge magnitude by the magnitude of the charge on a single electron.
SOLUTION The magnitude of the charge on an electron is e, so that the number N of electrons transferred is
7-REASONING AND SOLUTION The magnitude of the force of attraction between the charges is given by Coulomb's law (Equation 18.1): , where are the magnitudes of the charges and r is the separation of the charges. Let FA and FB represent the magnitudes of the forces between the charges when the separations are rA and rB=rA/9, respectively. Then
Therefore, we can conclude that .
10-
10.REASONING The drawing at the right shows the set-up. The force on the +q charge at the origin due to the other +q charge is given by Coulomb’s law (Equation 18.1), as is the force due to the +2q charge. These two forces point to the left, since each is repulsive. The sum of /the two is twice the force on the +q charge at the origin due to the other +q charge alone.
SOLUTION Applying Coulomb’s law, we have
Rearranging this result and solving for d give
We reject the negative root, because a negative value for d would locate the +2q charge to the left of the origin. Then, the two forces acting on the charge at the origin would have different directions, contrary to the statement of the problem. Therefore, the +2q charge is located at a position of .
16- REASONING According to Newton’s second law, the centripetal acceleration experienced by the orbiting electron is equal to the centripetal force divided by the electron’s mass. Recall from Section 5.3 that the centripetal force Fc is the name given to the net force required to keep an object on a circular path of radius r. For an electron orbiting about two protons, the centripetal force is provided almost exclusively by the electrostatic force of attraction between the electron and the protons. This force points toward the center of the circle and its magnitude is given by Coulomb’s law.
SOLUTIONThe magnitude ac of the centripetal acceleration is equal to the magnitude Fc of the centripetal force divided by the electron’s mass: (Equation 5.3). The centripetal force is provided almost entirely by the electrostatic force, so Fc = F, where F is the magnitude of the electrostatic force of attraction between the electron and the two protons, Thus, . The magnitude of the electrostatic force is given by Coulomb’s law, (Equation 18.1), where and are the magnitudes of the charges, r is the radius of the orbit, and . Substituting this expression for F into , and using m = 9.11 1031 kg for the mass of the electron, we find that
20- REASONING The kinetic energy of the orbiting electron is (Equation 6.2), where m and v are its mass and speed, respectively. We can obtain the speed by noting that the electron experiences a centripetal force whose magnitude Fc is given by (Equation 5.3), where r is the radius of the orbit. The centripetal force is provided almost exclusively by the electrostatic force of attraction F between the electron and the protons, so Fc = F. The electrostatic force points toward the center of the circle and its magnitude is given by Coulomb’s law as (Equation 18.1),where and are the magnitudes of the charges.
SOLUTIONThe kinetic energy of the electron is
(6.2)
Solving the centripetal-force expression, (Equation 5.3), for the speed v, and substituting the result into Equation 6.2 gives
(1)
The centripetal force is provided almost entirely by the electrostatic force, so Fc = F, where F is the magnitude of the electrostatic force of attraction between the electron and the three protons. This force is given by Coulomb’s law, (Equation 18.1). Substituting Coulomb’s law into Equation 1 yields
Setting and , we have
37- REASONING The electric field is given by Equation18.2 as the force F that acts on a test charge q0, divided by q0. Although the force is not known, the acceleration and mass of the charged object are given. Therefore, we can use Newton’s second law to determine the force as the mass times the acceleration and then determine the magnitude of the field directly from Equation18.2. The force has the same direction as the acceleration. The direction of the field, however, is in the direction opposite to that of the acceleration and force. This is because the object carries a negative charge, while the field has the same direction as the force acting on a positive test charge.
SOLUTION According to Equation18.2, the magnitude of the electric field is
According to Newton’s second law, the net force acting on an object of mass m and acceleration a is F=ma. Here, the net force is the electrostatic force F, since that force alone acts on the object. Thus, the magnitude of the electric field is
The direction of this field is opposite to the direction of the acceleration. Thus, the field points along the x axis.