We consider the scattering of a charged particle in a Coulomb potential. The force between a particle A and the scattering center B is:
\[
F = \frac{q\,Q}{4\pi\varepsilon_0\,r^2} = \frac{a}{r^2}
\]
where \(q\) is the charge of the particle and \(Q\) the charge of the scattering center.
In the center of mass frame the angular momentum is:
\[
L =\mu r^2 \frac{\mathrm d \varphi}{\mathrm dt} = \mu r v \sin\varphi\stackrel{(r\rightarrow-\infty)}{=} \mu v_0 b
\tag{1}\]
with the reduced mass \(\mu=m_1 m_2/(m_1+m_2)\). The particle is deflected by the force in \(y\)-direction:
\[
F_y = \frac{a\,\sin\varphi}{r^2} = \mu \frac{\mathrm d v_y}{\mathrm d t}
\]
With Equation 1 we get:
\[
\frac{\mathrm d v_y}{\mathrm d t} = \frac{a\,\sin\varphi}{\mu v_0 b} \frac{\mathrm d\varphi}{\mathrm d t}
\]
To get the total deflection \(\vartheta\) we need to integrate from \(\mathrm A(-\infty)\) to \(\mathrm A(+\infty)\):
\[
\begin{aligned}
\int\limits_{0}^{v_0 \sin\vartheta} \mathrm d v_y &=
\frac{a}{\mu v_0 b} \int\limits_{0}^{\pi - \vartheta} \sin \varphi \; \mathrm d\varphi\\
v_0 \sin \vartheta &= \frac{a}{\mu v_0 b} \left(1 + \cos\vartheta \right)
\end{aligned}
\]
Using \((1 + \cos\vartheta) / \sin\vartheta = 1/(\tan(\vartheta/2))\) we find:
\[
b = \frac{a}{\mu v_0^2} \frac{1}{\tan(\vartheta/2)}
\tag{2}\]
From this, we can now calculate the differential cross section
\[
\frac{\mathrm d\sigma}{\mathrm d\Omega} = \frac{b}{\sin\vartheta}\frac{\mathrm db}{\mathrm d\vartheta}
\]
Using \(\sin\vartheta=2\sin(\vartheta/2)\cos(\vartheta/2)\) we can rewrite Equation 2 as follows:
\[
\frac{b}{\sin\vartheta} = \frac{1}{2}\frac{a}{\mu v_0^2}\frac{1}{\sin^2(\vartheta/2)}
\]
The derivative of Equation 2 with respect to \(\vartheta\) gives:
\[
\frac{\mathrm db}{\mathrm d\vartheta} = \frac{1}{2}\frac{a}{\mu v_0^2}\frac{1}{\sin^2(\vartheta/2)}
\]
Thus, for the differential cross section we get:
\[
\frac{\mathrm d\sigma}{\mathrm d\Omega} = \frac{1}{4}\left(\frac{a}{\mu v_0^2}\right)^2\frac{1}{\sin^4(\vartheta/2)} = \frac{1}{4}\left(\frac{q\,Q}{4\pi\varepsilon_0\,\mu v_0^2}\right)^2\frac{1}{\sin^4(\vartheta/2)}
\]
which is the Rutherford scattering formula. This formula successfully described the experimental data except at very large angles (small impact parameters), where deviations indicated that nuclei, while much smaller than atoms (< \(10^{-14}\) m), are not point-like. This groundbreaking experiment established the nuclear model of the atom and provided the first evidence for nuclear structure.