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+### Statement
+
+$7.4.10.$ [Insert the problem statement]
+
+### Solution
+
+### Problem Statement
+
+$7.4.10^*.$ The velocities of three charged particles of mass $m$ are shown in the figure. The distance from each particle to the edge of the metallic dihedral angle is $d$. The charges of the first two particles, flying in opposite directions, are equal to $-q$, and the charge of the third particle is $q$. Determine the velocity of these particles at a large distance from each other.
+
+![For problem $7.4.10$|308x248, 50%](../../img/7.4.10/Снимок экрана 2026-08-09 210121.png)
+
+### Solution
+
+#### 1. Physical model and the method of electrostatic images
+In problems of this class, it is assumed by default that the metallic faces of the right angle extend to infinity. This allows us to neglect edge effects and consider the entire surface of the angle to be strictly equipotential with a zero potential ($\Phi = 0$, referenced to infinity).
+
+The metallic dihedral angle occupies exactly one quarter of space. Our three real charges fly in the remaining empty space. To account for the influence of the charges induced on the metal, we apply the method of electrostatic images: we mentally remove the metallic angle and replace it with such fictitious charges that the potential at the location of the former walls still remains zero.
+
+An elegant geometric trick is applied here: the three initial charges already form three vertices of a square with side $a = d\sqrt{2}$ and diagonal $2d$. To nullify the potential at the boundaries of the angle, it is sufficient to place just one fictitious charge $+q$ in the empty fourth vertex of the square (in the region where the metal was located).
+
+A configuration of four charges at the corners of a square with alternating signs ($-q, +q, -q, +q$) creates a strictly zero potential on its midlines (axes of symmetry passing through the midpoints of opposite sides). These midlines perfectly coincide with the boundaries of our metallic angle.
+
+According to the uniqueness theorem in electrostatics, to uniquely determine the field, it is sufficient to ensure the correct boundary condition ($\Phi = 0$ on the walls). Since the boundary is zeroed, the electric field only in the region of the real charges (in the three empty quadrants) will be absolutely identical to the field of the vacuum configuration. The fact that in the imaginary vacuum model the potential beyond the boundary (in the zone of the fictitious charge) is not equal to zero does not matter — in reality, there is the bulk of the conductor there, the electrostatic field inside which is strictly zero by definition.
+
+#### 2. Energy of the electric field
+The key difference between the real system and the vacuum "square" is that in reality, the electric field exists only in three free quadrants of space. Inside the ideal conductor (in the fourth quadrant), there is no macroscopic electrostatic field.
+
+Since the energy is localized directly in the space of the field, and the field configuration in the vacuum equivalent is completely symmetric across the four quadrants, the initial interaction energy of the real system $W_0$ is exactly $3/4$ of the total energy of the vacuum configuration $W_{\text{vac}}$:
+$$W_0 = \frac{3}{4} W_{\text{vac}}$$
+
+Let us calculate $W_{\text{vac}}$ for four charges. The energy is the sum of the interactions along the four sides of the square (charges of opposite signs, distance $d\sqrt{2}$) and the two diagonals (charges of the same sign, distance $2d$):
+$$W_{\text{vac}} = 4 \left( \frac{1}{4\pi\varepsilon_0} \frac{(+q)(-q)}{d\sqrt{2}} \right) + 2 \left( \frac{1}{4\pi\varepsilon_0} \frac{q^2}{2d} \right)$$
+$$W_{\text{vac}} = -\frac{4q^2}{4\pi\varepsilon_0 d\sqrt{2}} + \frac{2q^2}{4\pi\varepsilon_0 (2d)} = \frac{q^2}{4\pi\varepsilon_0 d} \left( 1 - 2\sqrt{2} \right)$$
+
+Then the initial electrical energy of our real system is:
+$$W_0 = \frac{3}{4} \frac{q^2}{4\pi\varepsilon_0 d} (1 - 2\sqrt{2}) = -\frac{3q^2(2\sqrt{2} - 1)}{16\pi\varepsilon_0 d}$$
+
+#### 3. Kinematics and the law of conservation of energy
+It is fundamentally important to note: the fictitious charge $+q$ has no mass; it is exclusively a mathematical tool for calculating the field. The kinematics of the system is described strictly for three real particles of mass $m$.
+
+Due to symmetry, the Coulomb forces acting on each of the charges are directed strictly radially towards the vertex of the angle. The particles will fly apart synchronously, maintaining the shape of an expanding square. At infinity, the potential energy of the field will become zero, and the velocities of all three particles will become equal to $v$.
+
+Let us write the law of conservation of energy:
+$$3 \frac{mv_0^2}{2} + W_0 = 3 \frac{mv^2}{2}$$
+$$3 \frac{mv_0^2}{2} - \frac{3q^2(2\sqrt{2} - 1)}{16\pi\varepsilon_0 d} = 3 \frac{mv^2}{2}$$
+
+Divide the equation by 3:
+$$\frac{mv_0^2}{2} - \frac{q^2(2\sqrt{2} - 1)}{16\pi\varepsilon_0 d} = \frac{mv^2}{2}$$
+$$v^2 = v_0^2 - \frac{q^2(2\sqrt{2} - 1)}{8\pi\varepsilon_0 m d}$$
+$$v = \sqrt{ v_0^2 - \frac{q^2(2\sqrt{2} - 1)}{8\pi\varepsilon_0 m d} }$$
+
+*Note: The separation of particles to infinity is possible provided that the initial kinetic energy is sufficient to overcome the Coulomb attraction of the system (the expression under the square root must be $\geqslant 0$).*
+
+#### Answer:
+$$v = \sqrt{ v_0^2 - \frac{q^2(2\sqrt{2} - 1)}{8\pi\varepsilon_0 m d} } \quad \text{for} \quad \frac{mv_0^2}{2} \geqslant \frac{q^2(2\sqrt{2} - 1)}{16\pi\varepsilon_0 d}$$
+
+### Alternative Solution 1
+
+The metallic surface of the dihedral angle is an equipotential surface with potential $\Phi=0$. To satisfy this condition, the influence of the charges induced on the metal can be replaced by a single fictitious charge $+q$, located in the fourth quadrant.
+
+At the location of the charge $-q$ in the first quadrant, the potential is created by: the real charge $+q$ from the second quadrant at a distance $d\sqrt{2}$, the real charge $-q$ from the third quadrant at a distance $2d$, and the fictitious charge $+q$ at a distance $d\sqrt{2}$:
+
+$$\Phi_{1}=\frac{1}{4\pi\varepsilon_{0}}\left( \frac{q}{d\sqrt{2}}-\frac{q}{2d}+\frac{q}{d\sqrt{2}} \right)=\frac{1}{4\pi\varepsilon_{0}}\left( \frac{q\sqrt{2}}{d}-\frac{q}{2d} \right)=\frac{q(2\sqrt{2}-1)}{8\pi\varepsilon_{0}d}$$
+
+At the location of the charge $+q$ in the second quadrant, the potential is created by: two real charges $-q$ at a distance $d\sqrt{2}$ each, and the fictitious charge $+q$ at a distance $2d$:
+
+$$\Phi_{2}=\frac{1}{4\pi\varepsilon_{0}}\left( -\frac{q}{d\sqrt{2}}-\frac{q}{d\sqrt{2}}+\frac{q}{2d} \right)=\frac{1}{4\pi\varepsilon_{0}}\left( -\frac{q\sqrt{2}}{d}+\frac{q}{2d} \right)=-\Phi_{1}$$
+
+At the location of the charge $-q$ in the third quadrant, by symmetry, the potential is exactly equal to the potential of the charge $-q$ from the first quadrant:
+
+$$\Phi_{3}=\Phi_{1}$$
+
+The interaction energy of the real charges, taking into account the charges induced on the metal, is calculated only for the three real charges. The fictitious charge is not included in the sum because it is outside the field integration region (inside the conductor), and its field only mathematically models the fields of the charges induced on the angle:
+
+$$W_{0}=\frac{1}{2}\sum_{i=1}^{3}q_{i}\Phi_{i}=\frac{1}{2}\left( q_{1}\Phi_{1}+q_{2}\Phi_{2}+q_{3}\Phi_{3} \right)$$
+
+here the numbering corresponds to the quadrant number. Let's substitute the signs of the charges and the signs of their potentials:
+
+$$W_{0}=\frac{1}{2}\left( (-q)\cdot(+\Phi_{1})+(+q)\cdot(-\Phi_{1})+(-q)\cdot(+\Phi_{1}) \right)=-\frac{3}{2}q\Phi_{1}=-\frac{3q^{2}(2\sqrt{2}-1)}{16\pi\varepsilon_{0}d}$$
+
+The law of conservation of energy for the three particles of mass $m$:
+
+$$3\cdot\frac{m\upsilon_{0}^{2}}{2}+W_{0}=3\cdot\frac{m\upsilon^{2}}{2}$$
+$$3\cdot\frac{m\upsilon_{0}^{2}}{2}-\frac{3q^{2}(2\sqrt{2}-1)}{16\pi\varepsilon_{0}d}=3\cdot\frac{m\upsilon^{2}}{2}$$
+
+From here:
+
+$$\upsilon=\upsilon_{0}\sqrt{1-\frac{q^{2}(2\sqrt{2}-1)}{8\pi\varepsilon_{0}m\upsilon_{0}^{2}d}}$$
+
+This solution is valid provided that the initial kinetic energy is sufficient to overcome the Coulomb attraction, that is, when:
+
+$$\frac{m\upsilon_{0}^{2}}{2}\ge \frac{q^{2}(2\sqrt{2}-1)}{16\pi\varepsilon_{0}d}$$
+
+
+### Alternative Solution 2
+
+The field method. In field theory [1,2,3], the electrostatic energy of charges is equal to the energy of the field of these charges in all space:
+
+$$
+W=\frac{\varepsilon\varepsilon_0}{2}\int E^2\,dV,
+$$
+
+where $E$ is the field created by these charges, and the integral is taken over all space.
+
+By symmetry, the infinite metal cuts out a quarter of the energy in all space from the energy that the four real charges would have if we replaced the image charge in the metal with a real one and removed the metal. Therefore, we write for the initial electrical energy of the system with the metal:
+
+$$
+W_0=\frac{3}{4}\left(2\frac{q^2}{4\pi\varepsilon_0(2d)}-4\frac{q^2}{4\pi\varepsilon_0\left(\sqrt2d\right)}+W_\text{self}\right),
+$$
+
+where $W_\text{self}$ is the sum of the self electrical energies of the four charges (we consider an isolated charge by itself to possess energy [1,2,3]).
+
+For the final electrical energy of the system:
+
+$$
+W=\frac{3}{4}\left(W_\text{self}\right),
+$$
+
+since there is no one charge out of the four equal ones.
+
+The law of conservation of energy for the system with the metal:
+
+$$
+3\cdot\frac{m\upsilon_{0}^{2}}{2}+W_{0}=3\cdot\frac{m\upsilon^{2}}{2}+W.
+$$
+
+From which:
+
+$$
+\upsilon=\upsilon_{0}\sqrt{1-\frac{q^{2}(2\sqrt{2}-1)}{8\pi\varepsilon_{0}m\upsilon_{0}^{2}d}}.
+$$
+
+This solution is valid provided that the initial kinetic energy is sufficient to overcome the Coulomb attraction, that is, when:
+
+$$
+\frac{m\upsilon_{0}^{2}}{2}\ge \frac{q^{2}(2\sqrt{2}-1)}{16\pi\varepsilon_{0}d}.
+$$
+
+*Note: we do not take into account the work of the field on the near-surface induced charges in the metal during their redistribution, because even when going deep into the metal they cover a negligibly small distance, meaning the work of the field on the near-surface charges is negligibly small. The field in the near-surface layer is perpendicular to the conductor (see also section 23, problem 7 in [1]).*
+
+#### References
+
+[1] Sivukhin D.V. *Electricity. General Course of Physics.* – 1983.
+
+[2] L. Landau and E. Lifshitz, *The Classical Theory of Fields*, Gostekhizdat, 1948, § 37.
+
+[3] *Course of lectures on general physics, Part III Electricity and Magnetism, Electrostatics*, 2nd ed. Kryukov S.P., AESC MSU.
+
+
+#### Answer
+
+$$\upsilon=\upsilon_{0}\sqrt{1-\frac{q^{2}(2\sqrt{2}-1)}{8\pi\varepsilon_{0}m\upsilon_{0}^{2}d}}$$
+
+for $$\frac{m\upsilon_{0}^{2}}{2}\ge \frac{q^{2}(2\sqrt{2}-1)}{16\pi\varepsilon_{0}d}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]