ADDITIONS TO WOLFSON PROBLEMS AND PROBLEMS FROM OTHER TEXTS JULY 14 - JULY 20 WOLFSON - CHAPTER 25 E3. This is the exercise accompanying Example 25-4 on page 639. 35. (b) Plot E(r) and V(r) from r = 0 to r = 4.0 m for a = 0.5 m, b = 0.9 m, c = 1.0 m and Q = 1.0 microC. Halliday and Resnick - 2nd Ed - CHAPTER 26 19. A positive charge per unit length lambda is distributed uniformly along a straight-line segment of length L. (a) Determine the potential (chosen to be zero at infinity) at a point P a distance y from one end of the charged segment and in line with it. (b) Use the result of (a) to compute the component of the electric field at P in the y-direction (along the line). (c) Determine the component of the electric field at P in a direction perpendicular to the straight line. - . P | | y | | - + Figure for H 26:19 | + | + | + the line of plus marks L + stands for the line of | + positive charge | + | + - + 20. On a thin rod of length L lying along the x-axis with one end at the origin (x=0), there is distributed a positive charge per unit length given by lambda = cx, where c is a constant. (a) Taking the electrostatic potential at infinity to be zero, find V at the point P on the y-axis. (b) Determine the vertical component E_y of the electric field at P from the result of part (a). (c) Why cannot E_x, the horizontal component of the electric field at P, be found using the result of part (a)? y axis | | Figure for H 26:20 P is on the | y axis, a . P distance y | from the | the line of plus marks origin | stands for the line of | positive charge | --------+++++++++++++++++++------ x axis (0,0)|<------ L ------> | | (P prefixes) HRW Problem Supplement #1 - CHAPTER 25 58. Consider a flat, nonconducting ring of outer radius R and inner radius r = 0.200R; the ring has a uniform charge per unit area of sigma. With V = 0 at infinity, find an expression for the electric potential at point P on the central axis of the ring, at a distance z = 2.00R from the center of the ring. (P prefixes) HRW Problem Supplement #1 - CHAPTER 26 53. A certain parallel-plate capacitor is filled with a dielectric for which Kappa = 5.5. The area of each plate is 0.034 m^2, and the plates are separated by 2.0 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 200 kN/C. What is the maximum energy that can be stored in the capacitor? 66. Two parallel-plate capacitors A and B are connected in parallel across a 600 V battery. Each plate has area 80.0 cm^2 and the plate separations are 3.0 mm. Capacitor A is filled with air; capacitor B is filled with a dielectric of dielectric constant Kappa = 2.60. Find the magnitude of the electric field within (a) the dielectric of capacitor B and (b) the air of capacitor A. What are the free charge densities sigma on the higher-potential plate of (c) capacitor A and (d) capacitor B? (e) What is the induced charge density sigma' on the surface of the dielectric which is nearest to the higher-potential plate of capacitor B. Halliday and Resnick - 2nd Ed - CHAPTER 27 7. Two metal objects, a saw and a wrench, are lying side-by-side on an non-conducting table; the two metal objects are not in contact with one another; they have net charges of +70 pC and -70 pC, and this results in a 20 V potential difference between them. (a) What is the capacitance of the system? (b) If the charges are changed to +200 pC and -200 pC, what does the capacitance become? (c) What does the potential difference become? 13. The figure for this problem is Fig. 26-29, except the width of the slab is now b instead of 0.6d. A slab of copper of thickness b is thrust into a parallel-plate capacitor of plate area A; it is exactly halfway between the plates. (a) What is the capacitance after the slab is introduced? (b) If a charge q is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted? (c) How much work is done on the slab as it is inserted? Is the slab sucked in or must it be pushed in? 18. Two parallel plates of area 100 cm^2 are given charges of equal magnitudes 0.89 microC but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.4 MV/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface. 43. You are asked to construct a capacitor having a capacitance near 1 nF and a breakdown potential in excess of 10000 V. You think of using the sides of a tall Pyrex drinking glass as a dielectric, lining the inside and outside curved surfaces with aluminum foil to act as the plates. The glass is 15 cm tall with an inner radius of 3.6 cm and an outer radius of 3.8 cm. What are the (a) capacitance and (b) breakdown potential of this capacitor? 45. The figure for this problem is Fig. 26-29, except that the slab is a dielectric with kappa = 2.61 instead of a conductor, and a 85.5 V battery is connected to the capacitor plates while the dielectric is being inserted. The area A = 115 cm^2, d = 1.24 cm, and the width of the slab is (0.629)d. Calculate (a) the capacitance, (b) the charge on the capacitor plates, (c) the electric field in the gap, and (d) the electric field in the slab, after the slab is in place. WOLFSON - CHAPTER 28 27. (b) in the 6.0 Ohm resistor? (c) in the 2.0 Ohm resistor? (d) in the 1.0 Ohm resistor? (e) What is the power output of the battery?