Homework 2

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Homework 2 66. On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 100 N/C. Compare the gravitational and electric forces on a small dust particle of mass 2.0×10−15g2.0×10−15g that carries a single electron charge. What is the acceleration (both magnitude and direction) of the dust particle? F= -(-1.602x10^-19C)(100N/C) =(1.602x10^-17N) F=-(2.0x10^18kg)(9.8m/s^2) =-1.96x10^17N F=1.602x10^-17N-1.96x10^-17N =-0.358x10^-17N A=-0.358x10^-17N/2.0x10^-18kg =-1.79m/s^2 75. (a) What is the electric field of an oxygen nucleus at a point that is 10−10m10−10m from the nucleus? (b) What is the force this electric field exerts on a second oxygen nucleus placed at that point? a. E=9 x 10^9 x 8 x 1.6 x 10^-19/(10^-10)^2 E=1.15 x 10^12N/C b. F=8 x 1.6 x 10^-19 x 1.15 x 10^12 F= 1.47 x 10^-6 N 80. A thin conducting plate 1.0 m on the side is given a charge of −2.0×10−6C−2.0×10−6C . An electron is placed 1.0 cm above the center of the plate. What is the acceleration of the electron? E=-2.0 x 10^-6C/m^2 =-1.13 x 10^3 N/C F=(-1.6 x 10^-19 C)(-1.13 x 10^5 N/C) =1.808 x 10^-14 N A=1.81 x 10^-14 N/9.1 x 10^-31 kg =1.986 x 10^-16 m/s^2 81. Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ=4.0×10−6C/m. E=4.0 x 10^-6C/m/(2(3.14)(8.85 x 10^-12C^2/N x m^2)(2.0m) = 35967N/C

94. A proton enters the uniform electric field produced by the two charged plates shown below. The magnitude of the electric field is 4.0×105N/C,4.0×105N/C, and the speed of the proton when it enters is 1.5×107m/s.1.5×107m/s. What distance d has the proton been deflected downward when it leaves the plates? D= (8.0 x 10^-9s)+1/2(-3.84 x 10^13 m/s)(8.0 x 10^-9 s)^2 =-1.23 x 10^-3 m 95. Shown below is a small sphere of mass 0.25 g that carries a charge of 9.0×10−10C.9.0×10−10C. The sphere is attached to one end of a very thin silk string 5.0 cm long. The other end of the string is attached to a large vertical conducting plate that has a charge density of 30×10−6C/m2.30×10−6C/m2. What is the angle that the string makes with the vertical? F=9x10^10 x 44 x 10^-6/8.85 x 10^-12 =44.8 x 10^-4N 98. From a distance of 10 cm, a proton is projected with a speed of v=4.0×106m/sv=4.0×106m/s directly at a large, positively charged plate whose charge density is σ=2.0×10−5C/m2.σ=2.0×10−5C/m2. (See below.) (a) Does the proton reach the plate? (b) If not, how far from the plate does it turn around? a. D=(4.0 x 10^6 m/s^2/2(1.07 x 10^14 m/s^2) =0.075m(10^2cm/1m) =7.5 cm b. X=10cm-7.5cm =2.5cm

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