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Drawing of the problem

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

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1 Answer 1

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When the particle q is approaching the other particle, the distance r1 is decreasing. The kinetic energy connected to the radious depends on the derivative of r1. At the minimum distance r1 is no longer decreasing and assumes his lower value. After that point r1 start increasing when the particle moves away. So at the beginning the derivative of r1 is negative, at the end is positive. If we assume that the derivative of r1 is a continuos function, it must be 0 when r1 is the minimum.

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  • $\begingroup$ 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? $\endgroup$ Commented Oct 30, 2021 at 12:52
  • $\begingroup$ If you read my comment carefully, I didn't write kinetic enery, but kinetic energy linked to the radious. You can always think the kinetic energy in polar coordinates as composed by two terms : one linked to the variation of the radious and one linked to the variation of the theta angle. I'm reffering to the first one. $\endgroup$ Commented Oct 30, 2021 at 12:55
  • $\begingroup$ @AndrianoDelVincio Oh, I think I got the point. We splitted the kinetic energy into two terms that are function of the distance between two particles. So it's not function of the actual velocity of particle anymore. I think I missed the point of using effective potential energy. Now I figured out the aim of using it. Thanks for your answer. $\endgroup$ Commented Oct 30, 2021 at 12:59

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