Fall ‘12 PHY 122 Homework Solutions #3

Chapter 22 Problem 38 (II) A very long solid nonconducting cylinder of radius R1 is uniformly charged with a charge density ρE. It is surrounded by a concentric cylindrical tube of inner radius R2 and outer radius R3 as shown in Fig. 22–36, and it too carries a uniform charge density ρE. Determine the electric field as a function of the distance R from the center of the cylinders for (a) 0 < R < R1(b) R1 < R < R2 (c) R2 < R < R3 and (d) R3 < R (e) If ρE = 15 μC/m3 and R1= ½ R2 = 1/3 R3 = 0.5cm, plot E as a function of R from R = 0 to R=20.0cm. Assume the cylinders are very long compared to R3. Chapter 22 Problem 41 A flat ring (inner radius R0, outer radius 4R0) is uniformly charged. In terms of the total charge Q, determine the electric field on the axis at points (a) 0.25R0 and (b) 75R0 from the center of the ring. [Hint: The ring can be replaced with two oppositely charged superposed disks.] Chapter 22 Problem 44 Suppose the density of charge between r1 and r0 of the hollow sphere of Problem 29 (Fig. 22–32) varies as ρe = ρ0 r1/r. Determine the electric field as a function of r for (a) 0 < r < r1 (b) r1 < r < r0 and (c) r > r0. (d) Plot E versus r from r = 0 to r = 2 r0. Chapter 23 Question 4 An electron is accelerated by a potential difference of, say, 0.10 V. How much greater would its final speed be if it is accelerated with four times as much voltage? Explain. Solution Assuming the electron starts from rest in both cases, the final speed will be twice as great. If the electron is accelerated through a potential difference that is four times as great, then its increase in kinetic energy will also be greater by a factor of four. Kinetic energy is proportional to the square of the speed, so the final speed will be greater by a factor of two.

Chapter 23 Question 12 Suppose the charged ring of Example 23–8 was not uniformly charged, so that the density of charge was twice as great near the top as near the bottom. Assuming the total charge Q is unchanged, would this affect the potential at point P on the axis (Fig. 23–14)? Would it affect the value of E at that point? Is there a discrepancy here? Explain. Solution The potential at point P would be unchanged. Each bit of positive charge will contribute an amount to the potential based on its charge and its distance from point P. Moving charges to different locations on the ring does not change their distance from P, and hence does not change their contributions to the potential at P. The value of the electric field will change. The electric field is the vector sum of all the contributions to the field from the individual charges. When the charge Q is distributed uniformly about the ring, the y-components of the field contributions cancel, leaving a net field in the x-direction. When the charge is not distributed uniformly, the y-components will not cancel, and the net field will have both x- and y-components, and will be larger than for the case of the uniform charge distribution. There is no discrepancy here, because electric potential is a scalar and electric field is a vector. Chapter 23 Question 15 A conducting sphere carries a charge Q and a second identical conducting sphere is neutral. The two are initially isolated, but then they are placed in contact. (a) What can you say about the potential of each when they are in contact? (b) Will charge flow from one to the other? If so, how much? (c) If the spheres do not have the same radius, how are your answers to parts (a) and (b) altered? Solution (a)Once the two spheres are placed in contact with each other, they effectively become one larger conductor. They will have the same potential because the potential everywhere on a conducting surface is constant. (b) Because the spheres are identical in size, an amount of charge Q/2 will flow from the initially charged sphere to the initially neutral sphere so that they will have equal charges. (c) Even if the spheres do not have the same radius, they will still be at the same potential once they are brought into contact because they still create one larger conductor. However, the amount of charge that flows will not be exactly equal to half the total charge. The larger sphere will end up with the larger charge.

Chapter 23 Problem 4 The work done by an external force to move a -9.10 μC charge from point a to point b is 7.00 x 10-4 J. If the charge was started from rest and had 2.10 x 10-4 J of kinetic energy when it reached point b, what must be the potential difference between a and b? Chapter 23 Problem 14 A 32-cm-diameter conducting sphere is charged to 680 V relative to V = 0 at r =infinity. (a) What is the surface charge density σ? (b) At what distance will the potential due to the sphere be only 25 V? Chapter 23 Problem 20 Repeat Problem 19 assuming the charge density ρE increases as the square of the distance from the center of the sphere, and ρE at the center. Chapter 23 Problem 24 A point charge Q creates an electric potential of +185V at a distance of 15 cm. What is Q (let V=0 at r = infinity)? Chapter 23 Problem 26 Two point charges, 3.4μC and -2.20μC are placed 5.0 cm apart on the x axis. At what points along the x axis is (a) the electric field zero and (b) the potential zero? Let V = 0 at r = infinity. Chapter 23 Problem 34 Three point charges are arranged at the corners of a square of side l as shown in Fig. 23–29. What is the potential at the fourth corner (point A), taking V=0 at a great distance?