Problem
There's two charged particles $q$ and $q\prime$. $q$ is moving in the electric field that is created by particle $q\prime$. Find the smallest distance ($r_1$) between two particles.
Short Explanation of My Question
This is a problem my instructor solved in lecture but I've got a question about the solution.
My instructor wrote the mechanical energy of particle $q$ when the distance between $q$ and $q\prime$ is $r_1$ (smallest distance) as below.
$$E=T(r_1)+U(r_1)$$
$$T(r_1)=\frac{1}{2}m\dot{r_1}^2=0$$
$$U(r_1)=\frac{J^2}{2mr_1^2}+\frac{k}{r}$$
Here, $U(r_1)$ is effective potential energy of $q$. What I can not understand is how particle $q$ has 0 kinetic energy. When I asked this question to my instructor she told me that distance between two particles, $r_1$, doesn't change so it's derivative is 0. Can you help me about this please?
Solution of My Instructor
At the beginning $q$ is very far from $q\prime$, so it's potential energy can be neglected. So the initial energy is:
$$E=\frac{1}{2}mv^2$$
Energy of $q$ at distance $r_1$:
$$E=T(r_1)+U(r_1)$$
$$T(r_1)=\frac{1}{2}m\dot{r_1}^2=0$$
$$U(r_1)=\frac{J^2}{2mr_1^2}+\frac{k}{r}$$
$$\vec{J}=\vec{r}\times\vec{p}=m\left(\vec{r}\times\vec{v}\right)$$
$$J=mrv\sin(\alpha)=mrv\left(\frac{b}{r}\right)=mvb$$
$$U(r_1)=\frac{mv^2b^2}{2r_1^2}+\frac{k}{r_1}=\frac{mv^2b^2+2r_1k}{2r_1^2}$$
$$E=\frac{mv^2b^2+2r_1k}{2r_1^2}$$
Angular momentum and energy is covserved due to the inverse square force. Using conservation of energy:
$$\frac{1}{2}mv^2=\frac{mv^2b^2+2r_1k}{2r_1^2}$$
$$mv^2r_1^2=mv^2b^2+2r_1k$$
$$r_1^2=b^2+\frac{2r_1k}{mv^2}, a=\frac{k}{mv^2}$$
$$r_1^2-2ar_1-b^2=0$$
$$(r_1)_{1,2}=\frac{2a \pm\sqrt{4a^2+4b^2}}{2}=a \pm\sqrt{a^2+b^2}$$
$(r_1)_1=a-\sqrt{a^2+b^2}$ is negative and is not a physical solution for this problem so smallest distance between two particles is given as $(r_1)_2=r_1=a+\sqrt{a^2+b^2}$.
