In the world gap, a $m$ mass, $q$ charged particle is placed at a distance $d$ from an infinitely large conducting plane. After releasing the particle, how long will it take to hit the plane? ( $d=1 \mathrm{~m}, q=1.6 \cdot 10^{-19} \mathrm{C}, m=1.67 \cdot 10^{-27} \mathrm{~kg}$. )
I. solution. The electric field formed between the infinite conducting plane and the point charge $q$ is as if we placed a so-called "image charge" of $-q$ at the same distance on the other side of the plane. Thus, a force of magnitude $F(x) = \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{(2 x)^{2}}$ acts on the charge at a distance $x$ from the plane. We need to determine how long it takes for a body of mass $m$ to travel a distance $d$ in this varying electric field.

Let's rephrase the problem! The force acting on the charge $q$ is the same as if it were approaching a fixed point charge of $q^{\prime}=q / 4$ at a distance $x$. We know that in such an electric field, where a force proportional to $1 / r^{2}$ acts on the body at a distance $r$ from the origin, the bodies move along conic sections. In this case, the body moves along an elliptical orbit because its total energy is negative. The Kepler laws apply to different elliptical orbits.

In the given problem, the orbit of the charge $q$ can be considered an elongated ellipse with a minor axis of zero length. Consider this degenerate elliptical orbit and a circular orbit of radius $d$. The third Kepler law applies to both of these elliptical orbits, which states that

$$
\frac{T_{1}^{2}}{T_{2}^{2}}=\frac{r_{1}^{3}}{r_{2}^{3}}
$$

where $T_{1}$ and $T_{2}$ are the respective orbital periods, and $r_{1}$ and $r_{2}$ are the average distances from the origin. Let $r_{2}=d$, then the orbital period on the circular orbit can be calculated from the following equation:

$$
\begin{gathered}
\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{4 d^{2}}=m d\left(\frac{2 \pi}{T_{2}}\right)^{2} \\
\text { since } \omega=\frac{2 \pi}{T_{2}}
\end{gathered}
$$

For the degenerate elliptical orbit, $r_{1}=d / 2$, so using (1) and (2),

$$
T_{1}=2 \pi \frac{\sqrt{2 \pi \varepsilon_{0} m d^{3}}}{q}
$$

The collision time we are looking for is $t=T_{1} / 2$. Numerically:

$$
t=5.98 \text{ s}
$$

II. solution. In the previous solution, we saw that the motion of the body is as if two bodies of the same mass and opposite charges were approaching each other. To describe the motion, we use the law of conservation of energy! The relative potential energy of the two bodies, when they are at a distance of $2 x$ from each other:

$$
V(x)=-\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{2 x}
$$

The change in the relative potential energy of the two bodies is equal to the kinetic energy gained:

$$
\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q^{2}}{2 x}-\frac{q^{2}}{2 d}\right)=2 \cdot \frac{1}{2} m v^{2}
$$

where we used the fact that the motion of the two bodies is symmetric, so their speeds are always the same and initially the distance between them is $2 d$. Since $v=d x / d t$, the above equation gives a differential equation. After separating the variables, we get the following equation:

$$
\frac{\sqrt{8 \pi \varepsilon_{0} \cdot m}}{q} \int_{d}^{0} \frac{x \cdot d}{d-x} d x=\int_{0}^{t} d t=t
$$

The definite integral on the left can be found in tables. The sought collision time is then given by:

$$
t=\pi \frac{\sqrt{2 \pi \varepsilon_{0} m d^{3}}}{q}
$$

Numerically: $t=5.98 \text{ s}$