91影视

Two -q charges revolve around +q charge.

One -q charge is in a circular orbit and the other -q is oscillate in an extreme ellipse.

Charge both circular and ellipse orbits have the same energy.

The electron whose orbit is in the shape of an extreme ellipse moves faster when it is close to the positive charge whereas it slows down when reaches the farther region of the orbit, from the central charge.

For the charge that is moving in a circular orbit, the potential energy is given by

$PE=-\frac{e^2}{4蟺{鈭�}_{0}R}$

The circular orbit gives,

$\frac{e^2}{4蟺{鈭�}_{0}R}=\frac{m{v}^2}{R}$

Rearranging the above equation for$m{v}^2$

$m{v}^2=\frac{e^2}{4蟺{鈭�}_{0}R}$

The kinetic energy is then given as-

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\frac{e^2}{4蟺{鈭�}_{0}R} \end{gathered}$$

The total energy is then given as-

$$\begin{gathered} E=P鈥�E+KE \\ =-\frac{1}{2}\frac{e^2}{4蟺{鈭�}_{0}R} \end{gathered}$$

For the extreme ellipse orbit, the max radius happens when potential energy equals total energy.

$$\begin{gathered} -\frac{e^2}{4蟺{鈭�}_{0}R}=-\frac{1}{2}\frac{e^2}{4蟺{鈭�}_{0}R} \\ r=2R \end{gathered}$$

Hence, proved. The radius can't be more than$r_{max}=2R$.

b)

In case of the circular orbit, the motion of the 鈥搎 charge is uniform about the same distance from the central charge. To keep it simple, though, we will approximate the orbital motion along the extreme ellipse to be a (1-dimensional) radial oscillation about the central charge.

As the -q charge executes its simple harmonic motion along the radial direction, its speed changes, so the time it spends in each part of the path accordingly.

The probability density of finding the oscillating charge at a distance r from the central charge (located at r = 0) takes the form

$P\left(r\right)=-\frac{2}{x\sqrt{{\displaystyle \frac{r_{max}}{2}}-r}}$

The probability of finding the oscillating charge in the region between $\frac{r_{max}}{2}$and $r_{max}$is the integral

$P\left(\frac{r_{max}}{2}<r<r_{max}\right)={鈭�}_{\frac{r_{max}}{2}}^{r_{max}}p\left(r\right)dr$

Substituting for P(r) and using the proper trigonometric substitution, we find

$$\begin{gathered} P\left(\frac{r_{max}}{2}<r<r_{max}\right)=\frac{2}{蟺}{鈭�}_{\frac{r_{max}}{2}}^{r_{max}}p\left(r\right)dr \\ =\frac{2}{蟺}{\sin }^{-1}{\overline{)\frac{r}{r_{max}}}}_{\frac{r_{max}}{2}}^{r_{max}} \\ =\frac{2}{蟺}\left(\frac{蟺}{2}-\frac{蟺}{6}\right)=\frac{2}{3} \end{gathered}$$

Therefore, the negative charge, in the circular orbit, spends all of its time at a fixed radial distance $R=r_{max}/2$from the central charge. it is clear that in the elliptical orbit, 鈥搎 charge will be, on average, farther from its central positive charge.