magnitude of that force? (b)What is that minimum magnitude? Earth\'s atmosphere
ID: 1571368 • Letter: M
Question
magnitude of that force? (b)What is that minimum magnitude? Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, each square meter of Earth's surface would intercept protons at the average rate of 1500 protons per second. What would be the electric current intercepted by the total surface area of the planet? Figure 21-34a show's charged particles 1 and 2 that are fixed in place on an x axis. Particle 1 has a charge with a magnitude of |q_1| = 8.00e. Particle 3 of charge q_3 = +8.00e is initially on the x axis near particle 2. Then particle 3 is gradually moved in the positive direction of the x axis. As a result, the magnitude of the net electrostatic force F_2, net on particle 2 due to particles 1 and 3 changes. Figure 21-34b gives the x component of that net force as a function of the position x of particle 3. The scale of the x axis is set by x_3 = 0.80 m. The plot has an asymptote of F_2, net = 1.5 X 10^-25 N as x rightarrow infinity. As a multiple of e and including the sign, what is the charge q_2 of particle 2?Explanation / Answer
We know that 1500 protons crosses 1 m^2 of the Earth's upper atmosphere, so we need to calculate the surface area of the earth. Now, the upper atmospher is many miles up, but it's only thin so I will just use the radius of the earth to calculate the total surface area of the atmospher presented to outer space where cosmic rays come from.
SA = 4 x pi x Re^2
SA = 4 x 3.142 x 6378000m
SA =5.1118 x 10^14 sq m.
With 1500 protons entering per sec then the total number entering the earth' atmospher is:
1500 x 5.1118 x 10^14
7.669 x 10^17
Total charge then = +1.6 x 10^-19 x 7.669 x 10^17
Total charge = 0.1227C
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