Q20P Question: Figure 21-30a聽shows a... [FREE SOLUTION]

Chapter 21: Q20P (page 626)

Question: Figure 21-30ashows an arrangement of three charged particles separated by distanced. ParticlesAandCare fixed on thex-axis, but particleBcan be moved along a circle centered on particleA. During the movement, a radial line betweenAandBmakes an angle $胃$ relative to the positive direction of thex-axis (Fig. 21-30b). The curves in Fig. 21-30cgive, for two situations, the magnitude $F_{n e t}$ of the net electrostatic force on particleAdue to the other particles. That net force is given as a function of angleuand as a multiple of a basic amount $F_{0}$ . For example on curve 1, at $胃 = 180 掳$ , we see that $F_{n e t} = 2 F_{0}$ [. (a) For the situation corresponding to curve 1, what is the ratio of the charge of particleCto that of particleB(including sign)? (b) For the situation corresponding to curve 2, what is that ratio?

Short Answer

Expert verified

a)For the situation on curve 1, the ratio of the charge of particle C to that of particle B is -4
b)For the situation on curve 2, the ratio of the charge of particle C to that of particle B is 16

Step by step solution

The given data

Particles A and C are fixed on the x-axis, but Particle B moves along a circle centered on particle A.
$O n c u r v e 1, a t 胃 = 180 掳, w e s e e t h a t F_{n e t} = 2 F_{0} .$

Understanding the concept of Coulomb’s law

Since the particles A and C are fixed on the x-axis, then the force of B on C is either parallel or anti-parallel. According to Coulomb's law, the force is inversely changing with the square of their separation distance. Hence, the force due to C should be one-fourth as much as the value of the force due to B on A, because C is twice as far from A. As the charges are not the same, there is a factor in the charge ratio. Thus, the force due to C is one-fourth of the charge ratio which is the multiple factors to the force due to B.

Formula:

The magnitude of the electrostatic force between any two particles,

$F_{1} = K (\frac{|q_{1}| |q_{2}| \cos 胃}{r^{2}}) (1)$

a) Calculation of the ratio of charges of C to B for curve 1

The maximum force is $2 F_{0}$ .and occurs when 胃 = 180掳 (B is to the left of A, while C is to the right of A). We choose the minus sign for the force due to C considering the direction of B鈥檚 force as positive, thus, we write the relation using the concept and equation (1) as: $(f o r, 未 = \frac{q_{B}}{q_{B}}$ (for,)

$2 F_{0} = (1 - \frac{1}{4} 尉) F_{0} 尉 = - 4$

One way to think of the minus sign choice is $\cos (180 掳) = - 1$ . This is certainly consistent with the minimum force ratio (zero) at $胃 = 0 掳 = 01$ since that would also imply:

$0 = (1 + \frac{1}{4} 尉) F_{0} 尉 = - 4$

Hence, for both the minimum and maximum cases of force, the value of the required ratio is -4

b) Calculation of the ratio of charges of C to B for curve 2

The ratio of maximum to minimum forces for the case of curve 2 is given as:
$\frac{1.25}{0.75} = \frac{5}{3}$

Thus, using the concept and equation (1), we get the relation as: (for,)

$\frac{5}{3} = \frac{1 + \frac{1}{4} 尉}{1 - \frac{1}{4}} 尉 = 16$

Of course, this could also be figured as illustrated in part (a), looking at the maximum force ratio by itself and solving, or looking at the minimum force ratiodata-custom-editor="chemistry" $\frac{3}{4}$ at 胃 = 180潞 and solving for $尉$ .

Hence, the value of the ratio is 16.