Section 1 Consider an object of mass B. 6.7 kg. Assume that it\'s made up of equ

Question

Section 1 Consider an object of mass B. 6.7 kg. Assume that it's made up of equal numbers of protons, neutrons, and electrons. Remember that protons and neutrons each have the same mass individually, namely 1.7 x10-27 kg; electrons have negligible mass by comparison; and that the charge of an individual proton (or electron) has magnitude 1.6 x 10-19 C. (a) How many protons does the object contain? Answer: 10 protons (b) Now imagine taking all the protons out of the object and stuffing them into a box. Also take all of the electrons out of the object and stuff them in a second box. How much charge is contained in each box? Answe r: x 10 C (c) Now imagine separating the two boxes - the one containing all the protons, the other containing all the electrons. Let's put them at a distance of C.9.6 m from one another. What will be the strength of the force of attraction between the two boxes? Answer: x 10 N (d) Suppose that you now want to take the two boxes and move them even farther apart, from a distance of C. 9.6 m to a distance of D. 11.8 m apart. How much work (that is, energy) would you have to expend to pull the boxes that much farther apart, fighting against the electrostatic attraction between them? Answer: x 10 J - (e) From this new distance of D. 11.8 m, if you were to release the clump of protons from their box, how much acceleration would the clump of protons undergo due to its attraction to the clump of electrons in the other box? Answer x 10 m/s2

Explanation / Answer

Let the box has 'n' protons, therefore it will have 'n' neutrons and 'n' electrons

mass of 'n' neutrons and protons=2*n*1.7*10-27 kg

total mass=6.7 kg

6.7=2*n*1.7*10-27

n=1.97*1027 protons

therefore number of nuetrons, electrons will be same

b) total charge in each box,Q = number of protons* charge on each proton

Q=1.97*1027*1.6*10*-19 C= 3.152*108 C

c) Force, F= qE= kQ2/r2= 9*109*(3.152*108)2 / (9.6)2

F=9.70225*1024 N

d) work done = change in PE

PE= kQ2/r ; initial PE= kQ2/r1 final PE =kQ2/r2

change in PE=kQ2*(1/r2 - 1/r1 ) {r2=11.8 m; r1 = 9.6 m}

WD= change in PE=-1.736*1025 J

e) acceleration,a= Force/mass

a=9.70*1024 / 1.7*10-27 m/s

a= 5.70*1051 m/s2

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