Chapter 18 Solutions Set Up: (a) The proton has charge and mass Let point a be at the negative plate and

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1 Chapter 18 Solutions *18.1. Set Up: Since the charge is positive the force on it is in the same direction as the electric field. Since the field is uniform the force is constant is upward is to the right, so is upward is upward, so is upward is at below the horizontal, so Reflect: The work is positive when the displacement has a component in the direction of the force it is negative when the displacement has a component opposite to the direction of the force. When the displacement is perpendicular to the force the work done is zero Set Up: (a) The proton has charge mass Let point a be at the negative plate point b be at the positive plate. The electric field is directed from the positive plate toward the negative plate. The force on the positively charged proton is in this direction. The proton moves in a direction opposite to the electric force, so the work done by the electric force is negative. Solve: * Set Up: For two oppositely charged parallel plates, where is the potential difference between the two plates, E is the uniform electric field between the plates, d is the separation of the plates. An electric field exerts a force on a charge placed in the field. An electron has charge mass Reflect: The electric field is the same at all points between the plates (away from the edges) so the acceleration would be the same at all points between the plates as it is for a point midway between the plates Set Up: For two oppositely charged sheets of charge, The positively charged sheet is the one at higher potential. The electric field is directed inward, toward the interior of the axon, since the outer surface of the membrane has positive charge points away from positive charge toward negative charge. Section 18.9 explores the effects of a material other than air between the plates. The outer surface has positive charge so it is at higher potential than the inner surface. * Set Up: From Example 18.4, with x rather than y as the distance from the negative plate, Solve: Reflect: says the potential increases linearly with x the graph of versus x should be a straight line, in agreement with the figure in the problem.

2 Set Up: (a) The direction of is always from high potential to low potential so point b is at higher potential. Apply Example 18.4 to relate to E. Solve: According to Example 18.4 we can calculate the magnitude of the electric field, E, to be Reflect: The electric force does negative work on a negative charge when the negative charge moves from high potential (point b) to low potential (point a) Set Up: Apply conservation of energy, Eq. (18.3). Use Eq. (18.10) to express U in terms of V. (a) Solve: Reflect: The electron gains kinetic energy when it moves to higher potential. Solve: Now Reflect: The electron loses kinetic energy when it moves to lower potential Set Up: Let a be when they are nm apart b when they are very far apart. A proton has charge mass As they move apart the protons have equal kinetic energies speeds. They have maximum speed when they are far apart all their initial electrical potential energy has been converted to kinetic energy. so so Their acceleration is largest when the force between them is largest this occurs at they are closest. when Set Up: From Problem 18.25, for a particle with charge of magnitude e. The speed of light is An electron has mass a proton has mass

3 From Problem 18.25, the kinetic energy the particle gains is so gives electrons protons the same kinetic energy. But the protons must be accelerated by a potential decrease whereas the electrons are accelerated by a potential increase. * Set Up: For a parallel-plate capacitor The surface charge density is (d) Reflect: If the plates are square, each side is about 18 cm in length. We could also calculate from * Set Up: The capacitors between b c are in parallel. This combination is in series with the 15 pf capacitor. For capacitors in parallel, so is in series with For capacitors in series, so Reflect: For capacitors in parallel the equivalent capacitance is larger than any of the individual capacitors. For capacitors in series the equivalent capacitance is smaller than any of the individual capacitors Set Up: For capacitors in series the voltage across the combination equals the sum of the voltages in the individual capacitors. For capacitors in parallel the voltage across the combination is the same as the voltage across each individual capacitor. Connect the capacitors in series so their voltages will add. where N is the number of capacitors in the series combination, since the capacitors are identical Set Up: In series, the charges on the capacitors are the same the sum of the potential differences across the capacitors is the applied potential difference. Let

4 The charge on each capacitor is The potential difference across the capacitor is 19.2 V. capacitor is 28.8 V the potential difference across the Set Up: For capacitors in series the voltages add the charges are the same; For capacitors in parallel the voltages are the same the charges add; The equivalent capacitance of the capacitors in parallel is When these two capacitors are replaced by their equivalent we get the network sketched in the figure below. The equivalent capacitance of these three capacitors in series is is the same as Q for each of the capacitors in the series combination shown in the figure below, so Q for each of the capacitors is Reflect: The voltages across each capacitor in the figure above are The sum of the voltages equals the applied voltage, apart from a small difference due to rounding. * Set Up: The charge across a capacitor charged to a voltage V is The energy stored in a charged capacitor is For capacitors in parallel we have for capacitors in series we have For the capacitors in parallel we have by the battery is For the capacitors in series we have The charge provided The charge provided by the battery is For the capacitors in parallel we have For the capacitors in series we have

5 Reflect: Since we also know the charge on each equivalent capacitor we could also use either or to calculate the energy supplied by the battery Set Up: The energy stored in a capacitor is The charge of an electron is The number of electrons with this magnitude of charge is To double the stored energy, halve the capacitance. To do this, either double the plate separation or halve the plate area. * Set Up: for a parallel plate capacitor; this equation applies whether or not a dielectric is present. per Reflect: The dielectric material increases the capacitance decreases the electric field that corresponds to a given potential difference Set Up: With the dielectric, before: after: The energy increased. Reflect: The power supply must put additional charge on the plates to maintain the same potential difference when the dielectric is inserted. so the stored energy increases. * Set Up: With air between the layers, The energy density in the electric field is The volume of a shell of thickness t average radius R is The volume of a solid sphere of radius R is With the dielectric present, has positive charge. For the cell, The outer wall of the cell is at higher potential, since it

6 The volume of the cell wall is The total electric field in the cell wall is (d) Set Up: Let a be the initial situation, where the alpha particle is very far from the gold nucleus has kinetic energy At a the gold nucleus has zero kinetic energy. Let b be at the distance of closest approach, when the distance between the two particles is Conservation of energy says The alpha particle has charge the gold nucleus has charge Solve: since at the distance of closest approach the alpha particle has momentarily come to rest. since is very large. Conservation of energy gives Set Up: where E is the total light energy output. The energy stored in the capacitor is Only 95% of the stored energy is available: The power output is 600 W, 95% of the original energy is converted, so so Reflect: For a given V, the stored energy increases linearly with C.