Mastering Pearson HW For Chapters 20 And 21 Physics

20.31 PART A: A +7.8 nC point charge and a -2.3 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between the two charges? 9.0x10^9 X 7.8×10^-9 11.6×10^-22 9.0X10^9 x 2.3×10^-9 (1.6×10^-292 = 274218.155 = 94921.875 E=E,+E2 3.5 x 10^5N/ 20.20 A platypus foraging for prey can detect an electric field as small as 0.002 N/C. PART A: To give an idea of sensitivity of the platypus   electric sense, how far from a +50nC point charge does the field have this magnitude? r= (9x10^9) (50x10^-9) 0.002 474.34 M E 20.26 A 25 nC charge experiences a 0.047 N electric force. PART A: What is the magnitude of the electric field at the position of this charge? E =F/2 →0.047N 25x10^-90 1.88×10^6W/L 20.28 A +20 nC charge is located at the origin. PART A: What is the strength of the electric field at the position (x,y) = (5.0cm, 0 cm)? (9×10^9) (20×10°C) = 12000WID 0.052 r = √x²+y² = D. DSM PART B: What is the strength of the electric field at the position (x,y) = (-5.0 cm, 5.0 cm)? (9x10^9) ( 20x10^-9() 0.072 r= √x2+y2 = 0.07m =36000N/L PART C: What is the strength of the electric field at the position (x,y) = (-5.0 cm, -5.0 cm)? r = sqrt((5.0)^2 + (-5.0)^2) = 7.07 cm (9 X 10^9) (20 X 10^-9) / (7.07 X 10^-2) = 3.6 X 10^-4 20.30 5.0 cm (+1.0 nC) ← 50 cm 5.0 m -1.0 nC PART A: What is the strength of the electric field at the position indicated by the dot in figure 1? 2(k q / r^2) sin θ (2) (9 X 10^9) (1 X 10^-9) / (5 X 10^-3) sin 45° = 2545.58 N/C 270° horizoutal 20.60 As shown in figure 1, a q=5.0nC charge sits at x=0 in a uniform E=1500 N/C electric field. PART A: At what point along the x-axis would a proton experience no net force? (9x10^9)(5x10^-9) ------------------ 1500 N/C x^2=0.03 x=0.173m) should be placed at the same point 20.63 A 0.15 g honeybee acquires a charge of 21 pC while flying. The electric field near the surface of the earth is typically 100 N/C, directed downward. PART A. What is the ratin of the electric force on the hee to the hee'e weiaht? 21 x 10^-12 ------------ = 1.42 x 10^6 0.15 x 10^-3 x 9.8 PART B: What electric field strength would allow the bee to hang suspended in air? 0.15 x 10^-3 x 9.8 -------------------- = 7 x 10^7 N/C 21 x 10^-12 20.72 An electric field E=(2.10x10^^5 N/C, right) causes the 2.0g ball in figure 1 to hang at an angle. PARTA: What is beta? θ = tan^-1 ((Fz / mg)) = (25x10^-9 x 2.10x10^5) / (2x10^-3 x 9.8) = 14.98° 21.18 Under normal circumstances, a separation of charges between the earth and the atmosphere produces an electric field that points downward. PART A: If you charge a balloon on your hair, it acquires a negative charge. Now consider and complete the description below for another situation. u = qv = -ev increases; decreases 21.4 In 1.0 second, a battery charger moves 0.62 C of charge from the negative terminal to the positive terminal of the 1.5 V AA battery. PART A: How much work does the charger do? work done = ΔV(ε) 1.5 (0.62) = (0.93) J 21.7 At one point in space, the electric potential energy of a 15 nC charge is 57 uJ PART A. What is the electrical potential at this noint? 57 x 10^-6 --------- = V x 15 x 10^9 15 x 10^-9 V = 3800V PART B: If a 25 nC charge were placed at this point, what would its electric potential energy be? 3800 x 25 x 10^-9 = 95 µJ 21.8 The scanning electron microscope image of bacterium shown in figure 1 was produced using a beam of electrons accelerated through a 30kV potential difference. PART A: What is the speed of the electrons? V = sqrt((2)(1.6 x 10^-19)(30 x 10^3)) / (9.11 x 10^-31) = 1.026 x 10^8 m/sec 21.11 An electron with an initial speed of 700,000m/s is brought to rest by an electric field. PART A: Did the electron move into a region of higher potential or lower potential? Because the electron is a negative charge and it slows down asit travels, it must be moving from a region of higherpatential to a region of lower potential. PART B: What was the potential difference that stooped the electron? ΔV = -mv²/2r -(9.11x10^-31 kg)(7x10^5 m/s)² / 2(1.60x10^-19 C) = -1.395 V PART C: What was the initial kinetic energy of the electron, in electron volts? (0.5)(9.11x10^-31)(7x7)x10^10 = 2.23x10^-19 / 1.6x10^-19 = 1.4 eV