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? 𝐸 1 = 9.0 × 1 0 9 × 7.8 × 1 0 − 9 ( 1.6 × 1 0 − 2 ) 2 = 274218.75   N/C E 1 ​ = (1.6×10 −2 ) 2 9.0×10 9 ×7.8×10 −9 ​ =274218.75N/C 𝐸 2 = 9.0 × 1 0 9 × 2.3 × 1 0 − 9 ( 1.6 × 1 0 − 2 ) 2 = 94921.875   N/C E 2 ​ = (1.6×10 −2 ) 2 9.0×10 9 ×2.3×10 −9 ​ =94921.875N/C 𝐸 = 𝐸 1 + 𝐸 2 = 3.5 × 1 0 5   N/C E=E 1 ​ +E 2 ​ =3.5×10 5 N/C 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 the sensitivity of the platypus electric sense, how far from a +50nC point charge does the field have this magnitude? 𝑟 = 𝑘 𝑞 𝐸 = ( 9 × 1 0 9 ) ( 50 × 1 0 − 9 ) 0.002 = 474.34   m r= E kq ​ ​ = 0.002 (9×10 9 )(50×10 −9 ) ​ ​ =474.34m 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 𝑞 = 0.047   N 25 × 1 0 − 9   C = 1.88 × 1 0 6   N/C E= q F ​ = 25×10 −9 C 0.047N ​ =1.88×10 6 N/C 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) = (5.0 cm, 0 cm)? 𝐸 = 9 × 1 0 9 × 20 × 1 0 − 9   C 0.0 5 2 = 72000   N/C E= 0.05 2 9×10 9 ×20×10 −9 C ​ =72000N/C 𝑟 = 𝑥 2 + 𝑦 2 = 0.05   m r= x 2 +y 2 ​ =0.05m PART B: What is the strength of the electric field at the position (x,y) = (-5.0 cm, 5.0 cm)? 𝐸 = 9 × 1 0 9 × 20 × 1 0 − 9   C 0.0 7 2 = 36700   N/C E= 0.07 2 9×10 9 ×20×10 −9 C ​ =36700N/C 𝑟 = 𝑥 2 + 𝑦 2 = 0.07   m r= x 2 +y 2 ​ =0.07m PART C: What is the strength of the electric field at the position (x,y) = (-5.0 cm, -5.0 cm)? 𝑟 = ( 5.0 ) 2 + ( − 5.0 ) 2 = 7.07   cm r= (5.0) 2 +(−5.0) 2 ​ =7.07cm 𝐸 = ( 9 × 1 0 9 ) ( 20 × 1 0 − 9 ) ( 7.07 × 1 0 − 2 ) 2 = 3.6 × 1 0 4   N/C E= (7.07×10 −2 ) 2 (9×10 9 )(20×10 −9 ) ​ =3.6×10 4 N/C 20.30 PART A: What is the strength of the electric field at the position indicated by the dot in figure 1? 𝐸 = 2 ( 𝑘 𝑞 𝑟 2 ) sin ⁡ 𝜃 E=2( r 2 kq ​ )sinθ 𝐸 = ( 2 ) ( 9 × 1 0 9 ) ( 1 × 1 0 − 9 ) sin ⁡ 4 5 ∘ ( 5 × 1 0 − 3 ) 2 = 2545.58   N/C E= (5×10 −3 ) 2 (2)(9×10 9 )(1×10 −9 )sin45 ∘ ​ =2545.58N/C PART B: What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line. 270° horizontal 270° horizontal 20.60 As shown in figure 1, a q = 5.0 nC 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? 𝐹 = 𝐸 ⋅ 𝑞 F=E⋅q 𝑥 2 = ( 9 × 1 0 9 ) ( 5 × 1 0 − 9 ) 1500   N/C x 2 = 1500N/C (9×10 9 )(5×10 −9 ) ​ 𝑥 = 0.03 = 0.173   m x= 0.03 ​ =0.173m PART B: At what point along the x-axis would the electron have no net force? 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 ratio of the electric force on the bee to the bee's weight? 21 × 1 0 − 12 0.15 × 1 0 − 3 × 9.8 = 1.42 × 1 0 − 6 0.15×10 −3 ×9.8 21×10 −12 ​ =1.42×10 −6 PART B: What electric field strength would allow the bee to hang suspended in air? 0.15 × 1 0 − 3 × 9.8 21 × 1 0 − 12 = 7 × 1 0 7   N/C 21×10 −12 0.15×10 −3 ×9.8 ​ =7×10 7 N/C PART C: What would be the necessary electric field direction for the bee to hang suspended in the air? upward upward 20.72 An electric field 𝐸 = ( 2.10 × 1 0 5   N/C , right ) E=(2.10×10 5 N/C,right) causes the 2.0 g ball in figure 1 to hang at an angle. PART A: What is beta? 𝜃 = tan ⁡ − 1 ( 𝐹 𝑒 𝑚 𝑔 ) θ=tan −1 ( mg F e ​ ​ ) 𝜃 = tan ⁡ − 1 ( 25 × 1 0 − 9 × 2.10 × 1 0 5 2 × 1 0 − 3 × 9.8 ) = 14.9 8 ∘ θ=tan −1 ( 2×10 −3 ×9.8 25×10 −9 ×2.10×10 5 ​ )=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 = 𝑞 𝑉 = − 𝑒 𝑉 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 = Δ 𝑉 ( 𝑞 ) Work done=ΔV(q) 1.5 × 0.62 = 0.93   J 1.5×0.62=0.93J PART B: Power output = 1.5 × 0.62 = 0.93   W Power output=1.5×0.62=0.93W 21.7 At one point in space, the electric potential energy of a 15 nC charge is 57 μJ. PART A: What is the electrical potential at this point? 57 × 1 0 − 6   J 15 × 1 0 − 9   C = 𝑉 15×10 −9 C 57×10 −6 J ​ =V 𝑉 = 57 × 1 0 − 6 15 × 1 0 − 9 = 3800   V V= 15×10 −9 57×10 −6 ​ =3800V PART B: If a 25 nC charge were placed at this point, what would its electric potential energy be? 3800   V × 25 × 1 0 − 9   C = 95   𝜇 J 3800V×25×10 −9 C=95μJ 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 = 2 ( 1.6 × 1 0 − 19   J ) ( 30 × 1 0 3   V ) 9.11 × 1 0 − 31   kg = 1.026 × 1 0 8   m/s v= 9.11×10 −31 kg 2(1.6×10 −19 J)(30×10 3 V) ​ ​ =1.026×10 8 m/s 21.11 An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. PART A: Did the electron move into a region of higher potential or lower potential? Answer: Because the electron is a negative charge and it slows down as it travels, it must be moving from a region of higher potential to a region of lower potential!