what is the magnitude of the gravitational force on one proton due to the other proton
Homework i
Chapter 16
Q9) The class of Coulomb's police force is very like to that for Newton's police of universal gravitation.
What are the differences between these 2 laws? Compare also gravitational mass and electric charge.
SOLUTION:The gravitational force is always attractive, and the magnitude of the force is: Gmiyard2/r2 . The electric force is attractive if the charges have contrary signs, and is repulsive if the charges have the same sign. The force magnitude is: kq1q2/rii . The gravitational and electric force are very similar in each having the changed distance- squared rule, and in having the magnitudes depending in a linear and symmetric fashion on properties of the 2 objects. The major difference is that possibility of repulsion for the electric force.
P2)Two charged fume particles exert a forcefulness of 4.2 ten 10-two N on each other. What will be the force if they are moved so they are only i eighth as far autonomously?
SOLUTION:
F=kq1qtwo/r2 = 4.2 x10-2 N If r -> (1/8)r, rtwo -> (one/64)rii and so,
F = 64(4.2 x x-2 N) = i.69 N
P6)What is the repulsive electrical force betwixt two protons in a nucleus that are five.0 x10-xv one thousand apart?
SOLUTION:
F = (9 x 109 Nm2/C2)(1.6 x 10-xix C)2 / (v.0 x10-15 grand)2 = ix.two N
P8)A person scuffing her feet on a wool carpet on a dry day accumulates a net charge of -60 µC.
How many excess electrons does this person get, and by how much does her mass increment?
SOLUTION:
Q = -60 µC, so the number of electrons is the total charge divided past the charge per electron:
Due northdue east = Q / (-east) = 60 µC/1.6x10-19 C
= 60 x10-half dozen/1.6x10-19 = 3.75 10 x14 backlog electrons.
The additional mass due to the excess electrons is the number of electrons multiplied by
the mass per electron:
Northeast( nine.xi x x-31kg) = three.42 x10-xvi kg
P11)Particles of charge +70, +48, and -fourscore µC are placed in a line. The center i is 0.35m from
each of the others. Calculate the cyberspace force on each due to the other ii (see Fig. 16-37 in text volume).
SOLUTION:
Allow Q1 = lxx µC , Qtwo = 48 µC, Q3 = -80 µC
Let the 10 axis lie along the charges. We volition calculate the x component of the forces (the other components vanish).
F1x = F12x + F13x = -k | Q1Qii | / (0.35 thousand)two + m | QaneQ3 | / (0.70 yard)2
F1x = [(ix.0 10 10nine Nm2/C2) / (0.70 m)ii] x (seventy µC) [-4 (48µC) + 80 µC]
= -144 N (force is to the left, indicated by negative value)
F2x = F21x + F23x = k | QaneQ2 | / (0.35 m)2 + one thousand | QiiQthree | / (0.35 1000)ii
F2x = [(nine.0 10 109 Nmii/C2) / (0.35 m)2] x (48 µC) [ (70µC) + lxxx µC]
= 529 North (force is to the correct, indicated by positive value)
F3x = F31x + F32x = -k | QaneQ3 | / (0.70 grand)ii - g | Q2Q3 | / (0.35 m)two
F3x = [(9.0 10 10ix Nmtwo/Cii) / (0.seventy m)ii] ten (lxxx µC) [ (-70µC) - 48 µC ten 4]
= 385 N (force is to the left, indicated by positive value)
P12)Three positive particles of charges eleven.0 µC are located at the corners of an equilateral triangle of side xv.0 cm (encounter Fig. 16-38 in text book). Calculate the magnitude and management of the net force on each particle.
SOLUTION:
Due to symmetry, each forcefulness is directed perpendicular to and away from the lines joining the other 2 charges. To calculate the magnitude, we only demand consider i example. For simplicity, the best case is, F1 , since it is apparent that F1x = 0.
F1y = F12y + F13y = | F12 | cos30 + | F13 | cos30 = 2 x |F12 | cos30
since the magnitude of the vector | F12 | = | Fthirteen |
| F1y | = two x [k (11 µC)2 / (0.15 m)2] x (iiiane/2/two) = 83.viii North so, | Fane | = | F2 | = | Fthree | = 83.eight N
P15)Compare the electric forcefulness holding the electron in orbit around the proton (r = 0.53 x 10-10m) in the hydrogen nucleus with the gravitational force between the aforementioned electron and proton. What is the ratio
of these two forces?
SOLUTION:
Electric force magnitude:
FE = (9 10 10nine Nm2/C2)(one.6 x ten-19 C)ii / (0.53 x 10-tenm)2
= 8.20 10 x-8 N
Gravitational forcefulness magnitude:
FM = (6.67 ten 10-11 Nm2/kg2)(9.0 10-31 kg) (one.67 x ten-27kg) / (0.53 x 10-xm)ii
= iii.61 x 10-47 N
The electrical force is much stronger, in the ratio FE / FG = ii.3 ten 1039
P18)In one model of the hydrogen cantlet, the electron revolves in a circular orbit around the proton with a speed of 1.1 x x6 m/due south. Decide the radius of the electron'southward orbit.
SOLUTION:
The centripetal force which keeps the electron in orbit is provided by the electrical force.
mv2/r = ke2 / rii and so, r = ke2 / mv2 = [(9 10 ten9 Nm2/C2)(1.6 x x-xix C)2] / [(ix.11 10 10-31 kg) x
(1.ane x 106 m/s)2] = two.09 x 10-10yard
BU CAS PY 106
This page maintained by Anna Skibinsky
askibins@buphy.bu.edu
Source: http://physics.bu.edu/py106/hwex/106hw1.html
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