Does the same point continue to be the neutral point when the system of charges start to move closer or away?

Consider two point charges $$+q_1$$ and $$+q_2$$ separated by a distance $$r$$. A charge $$q_0 ≠ -\frac{q_1q_2}{(\sqrt{q_1}+\sqrt{q_2})^2}$$ (n.b. the equality holds only when $$\vec{F}_{net,q_1} = \vec{F}_{net,q_2} = \vec{F}_{net,q_0} = \vec{0}$$. In that case no doubt the charges will be in equilibrium. So, I want to know that if I place any random charge(except that special magnitude) in the neutral point) is placed at a distance of $$\frac{r\sqrt{q_1}}{\sqrt{q_1}+\sqrt{q_2}}$$ from charge $$+q_1$$ before the charges $$q_1$$ and $$q_2$$ start to move. This distance is also called the neutral point. After the charge $$q_0$$ is placed, $$q_1$$ and $$q_2$$ will not be in equilibrium. So, will that same point continue to be the neutral point for the charges $$q_1$$ and $$q_2$$?

I learnt that

Force between two moving charges cannot be simply determined by Coulomb's Law.

Then I thought about conservation of momentum. Considering the charges $$q_1$$ , $$q_0$$ and $$q_2$$ in one system , the net external force is zero and the internal forces are conservative so the momentum must be conserved. This implies $$\vec{p}_i = \vec{p}_f$$ $$m_1\vec{v}_1 + m_0\vec{v}_0 + m_2\vec{v}_2 = \vec{0}$$ But still I am not able to draw any conclusions. Any method?

Update

As pointed out by @RichardMyers, according to Earnshaw's theorem a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction. If that is true then where is the fault in this :

$$F_1 = \frac{kq_1q_0}{r_0^2}$$ $$F_2 = \frac{kq_2 q_1}{r^2}$$ $$F_3 = \frac{kq_2 q_0}{(r-r_0)^2}$$ Now substituting $$q_0 = \frac{q_1q_2}{(\sqrt{q_1}+\sqrt{q_2})^2}$$ and $$r_0 = r \frac{\sqrt{q_1}}{\sqrt{q_1}+\sqrt{q_2}}$$ we get $$F_1 = F_2 = F_3$$ or $$\vec{F}_{net,q_1} = \vec{F}_{net,q_2} = \vec{F}_{net,q_0} = \vec{0}$$. So is it not a limitation of Earnshaw's theorem?

• I want to clarify, what is your initial charge configuration? Is it just the two positive charges that have nothing holding them in place? Jun 25, 2021 at 2:39
• Two charges $q_1$ and $q_2$ are separated by a distance $r$ in vacuum. Charge $q_0$ is placed at the neutral point before the charges $q_1$ and $q_2$ even start to move. Jun 25, 2021 at 3:30
• So am I to understand that the charge is placed where it feels no net force? I'm assuming by the statement about the magnitude of the third charge to mean that it is strong enough to pull the charges together against their repulsion (or let them move further out by not pulling them in enough)? That is it takes any value except that which would cause the system to not change? Jun 25, 2021 at 4:22
• It's worth pointing out that the momentum of a charge in an magnetic field (and moving charges will produce a magnetic field, is not given by $mv$. Jun 27, 2021 at 4:33
• Also, in any system of charges, $\vec F_{net,1}=\vec F_{net,2}=\vec F_{net,3}=0$ will never hold. This is known as Earnshaw's Theorem. Jun 27, 2021 at 4:36

The neutral point is the point where the resultant electric field is zero.

If we imagine a small test charge placed at the neutral point the force on it will be zero, but the other two charges will repel each other. Their motion then depends on their masses and in general this would cause the neutral point to change position.

John Hunter's solution is more elegant and concise; however, if you are interested also in the condition under which the point remains the neutral point and a more mathematical proof then here is a different approach:

In addition, I am not sure whether the questioner wanted the charge $$q_0$$ to impact the system or just as a test charge. If you wish $$q_0$$ to have no impact on the system then just set $$q_0=0$$ in the proceeding calculations.

Definitions

Because of the bulky equations, we will first define some notation:

$$\boldsymbol M\equiv\begin{bmatrix}m_0 & 0 & 0\\0 & m_1 & 0\\0 & 0 & m_2\end{bmatrix},\quad\quad\boldsymbol Q\equiv\begin{bmatrix}q_0 & 0 & 0\\0 & q_1 & 0\\0 & 0 & q_2\end{bmatrix},\quad\quad\vec{r}\equiv\begin{bmatrix}r_0\\r_1\\r_2\end{bmatrix},\quad\quad k\equiv\frac{1}{4\pi\epsilon\epsilon_0},\\\quad\\\text{and}\quad a\equiv\sqrt{q_1}+\sqrt{q_2}$$

where $$m_n$$, $$q_n$$ and $$r_n$$ are the mass, charge and position of the $$n^\text{th}$$ particle respectively. Finally, let $$r_{ij}\equiv r_i-r_j$$.

Equations of motion

Thus, the 1D equations of motion can concisely be written as:

$$\boldsymbol{M}\ddot{\vec{r}}=k\boldsymbol{Q}\begin{bmatrix}\frac{q_1}{r_{10}^2}+\frac{q_2}{r_{02}^2}\\\frac{q_2}{r_{21}^2}+\frac{q_0}{r_{10}^2}\\\frac{q_0}{r_{02}^2}+\frac{q_1}{r_{21}^2}\end{bmatrix}\tag{1}$$

Constraint

Now if we apply the condition that $$q_0$$ remains at the neutral point:

$$\left|r_{10}\right|=\frac{q_1}{a}\left|r_{21}\right|\quad\text{and}\quad\left|r_{02}\right|=\frac{q_2}{a}\left|r_{21}\right|$$

We can now use this to find the constraints on the parameters $$\boldsymbol{M}$$, $$\boldsymbol{Q}$$ and $$k$$ for which this is true.

Substituting these conditions into the left-hand side of (1) gives:

\begin{align}\boldsymbol{M}\ddot{\vec{r}}&=\frac{k}{r_{21}^2}\boldsymbol{Q}\underbrace{\begin{bmatrix}2a^2\\a^2+q_2\\a^2+q_1\end{bmatrix}}_{\equiv\vec{w}}\\\implies\ddot{\vec{r}}&=\frac{k}{r_{21}^2}\boldsymbol{M}^{-1}\boldsymbol{Q}\vec{w}\tag{2}\end{align}\\%to anyone who edits the extra line is to prevent clipping of bottom line in render

As $$\boldsymbol{M}$$ is diagonal:

$$\boldsymbol{M}^{-1}=\begin{bmatrix}\frac{1}{m_0} & 0 & 0\\0 & \frac{1}{m_1} & 0\\0 & 0 & \frac{1}{m_2}\end{bmatrix}\\%to anyone who edits the extra line is to prevent clipping of bottom of matrix in render$$

Thus equation (2) gives implies:

\begin{align}\ddot r_{21}&=\frac{k}{r_{21}}\left(\frac{q_2}{m_2}\left(a^2+q_1\right)-\frac{q_1}{m_1}\left(a^2+q_2\right)\right)\tag{3}\\\ddot r_{01}&=\frac{k}{r_{21}}\left(2\frac{q_0}{m_0}a^2-\frac{q_1}{m_1}\left(a^2+q_2\right)\right)\tag{4}\end{align}

However, we already know $$\left|r_{10}\right|=\frac{q_1}{a}\left|r_{21}\right|\implies\left|\ddot r_{10}\right|=\frac{q_1}{a}\left|\ddot r_{21}\right|$$

Thus, assuming the initial ordering of $$r_1 then $$\ddot r_{01}=\frac{q_1}{a}\ddot r_{21}$$ and substituting in equations (3) and (4) gives the constraint:

$$2\frac{q_0}{m_0}a^2-\frac{q_1}{m_1}\left(a^2+q_2\right)=\frac{q_1}{a}\left(\frac{q_2}{m_2}\left(a^2+q_1\right)-\frac{q_1}{m_1}\left(a^2+q_2\right)\right)\tag{5}$$

Similarly, if we had used $$r_{20}$$ we would have got:

$$\frac{q_2}{m_2}\left(a^2+q_1\right)-2\frac{q_0}{m_0}a^2=\frac{q_1}{a}\left(\frac{q_2}{m_2}\left(a^2+q_1\right)-\frac{q_1}{m_1}\left(a^2+q_2\right)\right)\tag{6}$$

Only if both (5) and (6) are satisfied then $$q_0$$ will remain at the neutral point. Note how both (5) and (6) only depend on charge and mass. Below I have plotted $$q_2$$ against $$q_1$$ for both equations (5) and (6) holding all other values constant for some non-zero parameters ($$m_n=1$$ for all $$n$$ and $$q_0=1.7$$) to demonstrate solutions do exist:

I believe the point about Earnshaw's theorem is that the resultant force on all three charges cannot be simultaneously zero but it can be zero for $$q_0$$ as you prove in the question.
For attractive forces, the charges should stay in a straight line. However, for repulsive forces, small deviations from the straight line will cause the charges to move away from the 1D line and so $$q_0$$ will no longer remain at the neutral point regardless of whether the parameters are a solution to (5) and (6). Unless the charges are somehow restricted to the only move along the 1D line.