a) In each situation, what is the direction of the net force on the central prot

Question

a) In each situation, what is the direction of the net force on the central proton due to the other particles?

b) In each situation, when the radius of the semicircle is Bohr radius, rBohr = 5.3

Figure 21-21, page 624 (pic posted below), shows four situations in which a central proton is partially surrounded by protons or electrons fixed in place along a half-circle. The angles theta theta are identical; the angles phi phi are also. a) In each situation, what is the direction of the net force on the central proton due to the other particles? b) In each situation, when the radius of the semicircle is Bohr radius, rBohr = 5.3 10^-11 m with theta theta = 30 degree and phi phi =45 degree, calculate the net force on the central proton due to the other particles? c) In each situation, calculate the net electric field on the central proton due to the other particles?

Explanation / Answer

in first case the direction of the net force is toward +ve x axis

in second case the direction of the net force is toward +ve x axis

in third case the direction of the net force is toward +ve x axis

in fourth case the direction of the net force is toward -ve x axis

the force due to one proton on central proton F

charge of proton = 1.6 * 10^-19 C

distance between protons = 5.3 * 10^-11 m

F = Ke * (charge of proton)^2 / (distance)^2

Ke = 8.98 * 10^9 N.m^2.C^-2

F = 8.98 * 10^9 * (1.6 * 10^-19)^2 / (5.3 * 10^-11)^2

F = 8.18 * 10^-8 N

in first case the net force on central proton will be = 2F * sin (theta) + 2F * cos (phi)

F1 = 2F * (sin (30) + cos (45))

F1 = 2 * 8.18 * 10^-8 * (sin (30) + cos (45))

F1 = 1.975 * 10^-7 N

in first case the net force on central proton will be = 1.975 * 10^-7 N

in second case the net force on central proton will be = same as in first case

in second case the net force on central proton will be = 1.975 * 10^-7 N

in third case the net force on central proton will be = -2F * sin (theta) - 2F * cos (phi)

F3 = 2F * (-sin (30) + cos (45))

F3 = 2 * 8.18 * 10^-8 * (-sin (30) + cos (45))

F3 = 3.88 * 10^-8 N

in third case the net force on central proton will be = 3.88 * 10^-8 N

in fourth case the net force on central proton will be = 2F * sin (theta) - 2F * cos (phi)

F1 = 2F * (sin (30) - cos (45))

F1 = 2 * 8.18 * 10^-8 * (sin (30) - cos (45))

F1 = -3.88 * 10^-8 N

in fourth case the net force on central proton will be = -3.88 * 10^-8 N

Electric field = force / charge

Electric field = F / charge of proton

in first case electric field = 1.975 * 10^-7 / 1.6 * 10^-19

in first case electric field = 1.23 * 10^12 N/C

in second case electric field = 1.23 * 10^12 N/C

in third case electric field = 3.88 * 10^-8 / 1.6 * 10^-19

in third case electric field = 2.45 * 10^11 N/C

in fourth casse electric field = -2.45 * 10^11 N/C

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