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 and

Solve: (a) is upward and is to the right, so and (b) is upward and is upward, so

(c) is upward and is at below the horizontal, so

Reflect: The work is positive when the displacement has a component in the direction of the force and 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.

18.2. Set Up: (a) The proton has charge and mass Let point a be at the negative plate and 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:  

*18.13. 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, and d is the separation of the plates. An electric field exerts a force on a charge placed in the field. An electron has charge

and mass

Solve: (a)    

(b)      

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.

18.14. Set Up: For two oppositely charged sheets of charge, The positively charged sheet is the one at higher potential.

Solve: (a) The electric field is directed inward, toward the interior of

the axon, since the outer surface of the membrane has positive charge and points away from positive charge and toward negative charge. Section 18.9 explores the effects of a material other than air between the plates. (b) The outer surface has positive charge so it is at higher potential than the inner surface.  

*18.17. Set Up: From Example 18.4, with and x rather than y as the distance from the negative plate,

Solve:  

Reflect: says the potential increases linearly with x and the graph of versus x should be a straight line, in agreement with the figure in the problem.

18.18. Set Up: (a) The direction of is always from high potential to low potential so point b is at higher potential. (b) 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

(c)   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).

18.20. 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. (b) Solve: Now

Reflect: The electron loses kinetic energy when it moves to lower potential.  

18.24. Set Up: Let a be when they are 0.750 nm apart and b when they are very far apart. A proton has charge and mass As they move apart the protons have equal kinetic energies and speeds.

Solve: (a) They have maximum speed when they are far apart and all their initial electrical potential energy has been converted to kinetic energy. and so

so and

(b) Their acceleration is largest when the force between them is largest and this occurs at when they are closest.

18.26. Set Up: From Problem 18.25, for a particle with charge of magnitude e. The speed of light is An electron has mass and a proton has mass

Solve: (a)      

(b) From Problem 18.25, the kinetic energy the particle gains is so gives electrons and protons the same kinetic energy. But the protons must be accelerated by a potential decrease whereas the electrons are accelerated by a potential increase.

(c)    

 

*18.39. Set Up: For a parallel-plate capacitor The surface charge density is

Solve: (a)    

(b)    

(c)    

(d)    

Reflect: If the plates are square, each side is about 18 cm in length. We could also calculate from

 

*18.47. Set Up: The capacitors between b and c are in parallel. This combination is in series with the 15 pF capacitor. and Solve: (a) For capacitors in parallel, so

(b) is in series with For capacitors in series, so

and

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.  

18.48. 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. Solve: (a) Connect the capacitors in series so their voltages will add. (b) where N is the number of capacitors in the series combination, since the capacitors are identical.  

18.52. Set Up: In series, the charges on the capacitors are the same and the sum of the potential differences

across the capacitors is the applied potential difference. Let and

Solve: (a)      

The charge on each capacitor is

(b)      

The potential difference across the capacitor is 28.8 V and the potential difference across the capacitor is 19.2 V.  

18.55. Set Up: For capacitors in series the voltages add and the charges are the same; For

capacitors in parallel the voltages are the same and the charges add;

Solve: (a) The equivalent capacitance of the and 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 (b)    (c) 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

and The sum of the voltages equals the

applied voltage, apart from a small difference due to rounding.

*18.57. 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 and for

capacitors in series we have

Solve: (a) For the capacitors in parallel we have The charge provided by the battery is

(b) For the capacitors in series we have The charge provided by the

battery is

(c) For the capacitors in parallel we have For the capacitors in series

we have

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.

18.62. Set Up: The energy stored in a capacitor is The charge of an electron is

Solve: (a) The number of electrons with this

magnitude of charge is

(b) To double the stored energy, halve the capacitance. To do this, either double the plate separation or halve the plate area.

*18.69. Set Up: for a parallel plate capacitor; this

equation applies whether or not a dielectric is present.

Solve: (a) per

(b)    

Reflect: The dielectric material increases the capacitance and decreases the electric field that corresponds to a given potential difference.

18.71. Set Up:  Solve: (a) With the dielectric,

before:

after:

(b) 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.

*18.77. Set Up: With air between the layers, and The energy density in the electric

field is The volume of a shell of thickness t and average radius R is The volume of a solid

sphere of radius R is With the dielectric present, and

Solve: (a)    

(b) The outer wall of the cell is at higher potential, since it has positive charge. (c) For the cell, and

The volume of the cell wall is

The total electric field in the cell wall is

(d) and

18.78. Set Up: Let a be the initial situation, where the alpha particle is very far from the gold nucleus and 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 and the

gold nucleus has charge

Solve: and since at the distance of closest approach the alpha particle has momentarily come to rest. since is very large.

Conservation of energy gives

18.84. 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:

Solve: (a) The power output is 600 W, and 95% of the original energy is converted, so

(b) so

Reflect: For a given V, the stored energy increases linearly with C.