Question

Figure 21-35 shows electrons 1 and 2 on an xaxis and charged ions 3 and 4 of identical charge$-q$and at identical angles$\mathit{\theta }$. Electron 2 is free to move; the other three particles are fixed in place at horizontal distances Rfrom electron 2 and are intended to hold electron 2 in place. For physically possible values

of $\mathit{q}\mathbf{<}\mathbf{5}\mathit{e}$, what are the (a) smallest, (b) second smallest, and (c) third smallest values offor which electron 2 is held in place?

Short answer

1. The smallest possible physical value of for which electron 2 is held in place is $37.5\left(0.654rad\right)$.
2. The second smallest possible physical value offor which electron 2 is held in place is $50.{95}^0\left(or0.889rad\right)$.
3. The third smallest possible physical value offor which electron 2 is held in place is $56.6\left(or0.988rad\right)$.

Step-by-step solution

Stating the given data 

1. Two electrons and two charged ions of identical charge is -q.
2. Electrons 1 and 2 are on x-axis, while ions 3 and 4 are at identical angles $\theta$.
3. Physically possible values are: $q\le 5e$.

Understanding the concept of Coulomb’s law

Using the basic concept of symmetry in Coulomb's law, we can get the possible angles of the charged ions. Using the concept of the graph, the electrons' positions can be determined. Again, using the concept of quantization, we can get the physically possible values of angles.

Formulae:

The magnitude of the electrostatic force between any two particles,

$F=k\frac{\left|q_1\right|\left|q_2\right|\cos \theta }{r^2}$ (i)

The angle between two position values, $\tan \theta =\left|\frac{y}{x}\right|$ (ii)

a) Calculation of the smallest possible value of θ

Let be the vertical distance from the coordinate origin to $q_3=-q$and $q_4=-q$on the +y axis, where the symbol q is assumed to be a positive value. Similarly, d is the (positive) distance from the origin with $q_4=-q$on the -y axis. If we take each angle θ in the figure to be positive, then we have $\tan \theta =d/r$

and

$\cos \theta =\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$r$}\right.$

whereis the dashed line distance shown in the figure.

From symmetry, we see that there is no net force in the vertical direction on $q_2=-e$sitting at a distance R to the left of the coordinate origin. Thus, the net x force caused by $q_{3}and{q}_4$and on the y-axis will have the magnitude, using equation (i), of

$$\begin{gathered} F_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$net$}\right.}=\frac{2qe}{4\mathrm{\pi }{\in }_0{\mathrm{r}}^2}\cos \theta \\ {F}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$net$}\right.}=\frac{2qe\cos \theta }{4\mathrm{\pi }{\in }_0{\left(\mathrm{R}/\cos \mathrm{\theta }\right)}^2}-\frac{2qe{\cos }^2\theta }{4\mathrm{\pi }{\in }_0{\mathrm{R}}^2}\left(fromequation\left(a\right)\right) \end{gathered}$$

Consequently, to achieve a zero net force along the x axis, the above expression must be equal to the magnitude of the repulsive force exerted on$q_2$by $q_1$.

Thus, this relates the net repulsive force $F_{2net}$with the force $F_{1net}$using

equation (i) as follows:

$F_{2net}=\frac{e^2}{4\mathrm{\pi }{\in }_0{\mathrm{R}}^2}$

$$\begin{gathered} F_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$net$}\right.}=\frac{e^2}{4\mathrm{\pi }{\in }_0{\mathrm{R}}^2}\left(fornetzeroforce\right) \\ \frac{2qe{\cos }^3\theta }{4\mathrm{\pi }{\in }_0{R}^2}=\frac{e^2}{4\mathrm{\pi }{\in }_0{\mathrm{R}}^2} \\ q=\frac{e}{2{\cos }^3\theta }----------\left(1\right) \end{gathered}$$

Now, plotting $\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$e$}\right.$as a function of the angle (in degrees)

According to the graph, $\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$e<5isfor\theta <{60}^0$}\right.,roughly$.

But, substituting equation (b) in our requirement $q\le 5e$, we can get

$$\begin{gathered} \frac{e}{2{\cos }^2\theta }\le 5e \\ \frac{1}{{\left(10\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\le \cos \theta \\ \theta \le 62.{34}^o \end{gathered}$$

Now, for “physically possible values’, it is reasonable to suppose that only positive-integer-multiple values of are allowed for. Comparing with equation (b), we get the value of integral multiple as:

$$\begin{gathered} n=\frac{1}{2{\cos }^2\theta } \\ \theta ={\cos }^{-1}\left\{{\left(\frac{1}{2n}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right\} \end{gathered}$$

where$n=1,2,....5$

Thus, the smallest possible value of angle is${\theta }_1=37.5^o\left(or0.654rad\right)$

Hence, the required value of the smallest angle is $37.5^o\left(or0.654rad\right)$

b) Calculation of the second smallest possible value of θ

Now, using the equation (b) from part (a) calculations, we can get the value of the second smallest angle as. ${\theta }_2=50.{95}^o\left(or0.889rad\right)$

Hence, the required value of the angle is$50.{95}^o\left(or0.889rad\right)$.

c) Calculation of the third smallest possible value of θ

Now, using the equation (b) from part (a) calculations, we can get the value of the third smallest angle as ${\theta }_3=56.6\left(or0.988rad\right)$.

Hence, the required value of the angle is $56.6^o\left(or0.988rad\right)$.