A Deflected Charge

Let

• Initial momentum $= \vec p_i$
• Final momentum $= \vec p_f$
• Change in momentum $= \Delta \vec p$
• Angle made by $\Delta \vec p$ with $\vec p_i$ , and due to symmetry, with $\vec p_f$ be $\varphi$
• Angle made by instantaneous force vector, $\vec F$ with $\Delta \vec p$ be $\psi$
• $k = \dfrac1{4\pi\varepsilon_0}$
• Kinetic Energy, $T = \frac12mv^2$
• Distance $D = \dfrac{kqQ}{T}$

So, we can say $p^2 = 2mT$

Due to symmetry, $\left|\vec p_i\right|=\left|\vec p_f\right|= p\text{ (Say)}$

Since, the path is symmetric, we need only take component of force along the change in momentum to get its magnitude.

$F_r = \dfrac{kqQ}{r^2}\\\begin{aligned}\left|\Delta\vec p\right| &= \int_{-\infty}^\infty F\cos\psi ~dt\\ &=\int_{-\infty}^\infty\dfrac{kqQ}{r^2}\cos\psi~dt\end{aligned}$

Conserving angular momentum, $pb =mr^2\omega=mr^2\dot\psi$ So, $dt = \dfrac{d\psi}{\dot\psi}=\dfrac{mr^2}{pb}d\psi$

From basic geometry, $2\varphi + \theta = \pi\\\varphi =\dfrac\pi2-\dfrac\theta2$

Hence, $\begin{aligned}\left|\Delta\vec p\right| &=\int_{-\infty}^\infty\dfrac{kqQ}{r^2}\cos\psi~dt\\ &=\int_{-\varphi}^\varphi\dfrac{kqQ}{r^2}\cos\psi\dfrac{mr^2}{pb}~d\psi\\ &=\int_{-\varphi}^\varphi\dfrac{kqQm}{pb}\cos\psi~ d\psi\\ &=\dfrac{2kqQm}{pb}\sin\varphi\\&=\dfrac{2kqQm}{pb}\cos\dfrac\theta2\end{aligned}$

Using sine rule, $\dfrac{\left|\vec p_i\right|}{\sin\varphi} = \dfrac{\left|\Delta\vec p\right|}{\sin\theta}\\ \Rightarrow\left|\Delta\vec p\right|=2p\sin \dfrac\theta2$

$\Rightarrow 2p\sin \dfrac\theta2 = \dfrac{2kqQm}{pb}\cos\dfrac\theta2\\ \begin{aligned} \Rightarrow\tan\dfrac\theta2 &= \dfrac{kqQm}{p^2b}\\ &=\dfrac{kqQ}{2Tb}\\&=\dfrac D{2b}\\&=\dfrac{kqQ}{mv^2b}\end{aligned}$

$\large\therefore \boxed{\begin{aligned}\theta &= 2\tan^{-1}\left(\dfrac D{2b}\right)\\&=2\tan^{-1}\left(\dfrac{qQ}{4\pi\varepsilon_0mv^2b}\right)\end{aligned}}$

Just solve the equation of motion to find the trajectory of the particle. The equation of motion of the scattered particle is $m\mathbf{\ddot{r}} \; = \; \dfrac{qQ}{4\pi \epsilon_0r^2}\mathbf{\hat{r}}$ where $\mathbf{r}$ is the position vector of the particle, and $\mathbf{\hat{r}}$ is a unit vector in the direction of $\mathbf{r}$ . Then $\mathbf{r} \times \mathbf{\ddot{r}} \,=\, \mathbf{0}$ , and so $\mathbf{h} \,=\, \mathbf{r} \times \mathbf{\dot{r}}$ must be constant. Introduce a fixed right-handed orthonormal triad of unit vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$ so that $\mathbf{h} \,=\, h\mathbf{k}$ where $h > 0$ and where $\mathbf{i}$ points along the direction from which the incident particle is coming. Since $\mathbf{r} \cdot \mathbf{h} = 0$ , we deduce that the particle moves entirely in the plane defined by the vectors $\mathbf{i}, \mathbf{j}$ , so we can define planar polar coordinates $r,\theta$ such that $\mathbf{\hat{r}} \; = \; r\cos\theta\mathbf{i} + r\sin\theta \mathbf{j} \;.$ Standard calculus shows us that $r^2\dot{\theta} = h$ and that the differential equation becomes $m\big(\ddot{r} - h^2r^{-3}\big) \; = \; \dfrac{qQ}{4\pi \varepsilon_0m}r^{-2} \; = \; kr^{-2}$ If we define $u = r^{-1}$ , then $\dot{r} \; =\; -h\dfrac{du}{d\theta} \qquad \ddot{r} \; = \; -h^2u^2 \dfrac{d^2u}{d\theta^2}$ and so the equation becomes $\dfrac{d^2u}{d\theta^2} + u \; = \; -\dfrac{k}{h^2m}$ which has the general solution $u \;= \; A\cos\theta + B\sin\theta - \dfrac{k}{h^2m} \;.$ Since the particle is at infinity with incoming speed $v$ with impact parameter $b$ at time $t=0$ , we see that $h = vb$ and moreover that $u=0$ and $\dfrac{du}{d\theta} = b^{-1}$ when $\theta=0$ . Thus $u \; = \; \dfrac{k}{h^2m}\cos\theta + \dfrac{1}{b}\sin\theta - \dfrac{k}{h^2m}$ . If we put $R \; = \; \sqrt{\dfrac{k^2}{h^4m^2} + \dfrac{1}{b^2}} \qquad \alpha \; = \; \tan^{-1}\dfrac{h^2m}{kb} \; = \; \tan^{-1}\dfrac{v^2b m}{k} \; = \; \tan^{-1}\left(\dfrac{4\pi \epsilon_0 mbv^2}{qQ}\right)$ then we have $u \; = \; R\big[ \cos(\theta - \alpha) - \cos(\alpha) \big]$ and hence $u = 0$ when $\theta-\alpha \,=\, \pm\alpha$ , and so when $\theta = 0\,,\,2\alpha$ .

Thus the scattering angle is $\pi - 2\alpha$ radians, which evaluates to $101.4^\circ$ for the given constants of this problem.

Using the Energy , Angular momentum conservation we get $r_m$

so $L=mv_0 b$

hence $E=\frac{1}{2}mv_0^2 +\frac{k_eQ}{r_0}$

$\frac{1}{2}mv^2 +\frac{k_eQ}{r_m}=\frac{1}{2}mv_0^2 +\frac{k_eQ}{r_0}$

$r_0 \rightarrow \infty$ so

$\frac{1}{2}mv^2 +\frac{k_eQ}{r_m}=\frac{1}{2}mv_0^2$

$\frac{1}{2}mv^2 +\frac{k_eQ}{r_m}-\frac{1}{2}mv_0^2 =0$

using Angular momentum conservation we yield that

$L=mvr_m sin(\pi/2) =mvr_m \rightarrow mvr_m=mv_0b \rightarrow v=v_0b/r_m$

so

$\frac{1}{2}m(v_0\frac{b}{r_m})^2 +\frac{k_eQ}{r_m}-\frac{1}{2}mv_0^2 =0$

so we multiply bothsides with $r+m^2$ so

$\frac{1}{2}m(v_0b)^2 +k_eQr_m-\frac{1}{2}mv_0^2 r_m^2=0$

so $r_m= \frac{-b -\sqrt{(k_eQ)^2 -4\frac{1}{2}mv_0^2 * \frac{1}{2}m(v_0b)^2 }}{mv_0^2}$

Now the path represent an hyperbola , $a=r_m -\frac{b}{cos\theta /2}, c=\frac{b}{cos\theta/2}$

we know that we can express a hyperbola in polarcoordinates with

$r=\frac{r_m}{1-\epsilon cos\theta }$ when r gets to infinity $1-\epsilon cos\theta =0$

$1=\epsilon cos\theta$

$1=\frac{c}{a}cos\theta$

solving the Trigonometric equation we yield

$\theta =1.77rad$