# Finding the smallest distance between two charged particles

## 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 $$(r_1)_2=r_1=a+\sqrt{a^2+b^2}$$.

• Okay, that seems valid but the particle $q$ had an initial velocity $v$. Electric field can not stop this particle so it should always have a kinetic energy. Am I false? Commented Oct 30, 2021 at 12:52