Suppose n charges are put in an arbitrary electrostatic configuration and a small test charge is placed at a null point (i.e., where $\vec{E}=0$ ) of the configuration. The task is to show that the equilibrium of the test charge is necessarily unstable. According to my textbook, if the equilibrium was stable, then the test charge when displaced a little will experience a restoring force towards the null point, this implies that when a Gaussian surface is considered around the null point, there will be a net electric flux in or out of the surface which implies that there is a net charge enclosed by the surface which contradicts that no charge was present inside the surface.

Now I tried to see it mathematically, so what I did is to consider field at null point and at the displaced location.

$\displaystyle{ \vec{E}_{null point} = \vec{0} = \frac{1}{4 \pi \epsilon_{o}} \sum ^{n} _{i}{\frac{q_{i}(\vec{r_{o}}-\vec{r_{i}})}{|\vec{r_{o}}-\vec{r_{i}}|^{3}}} }$

Now if I consider electrostatic field at a point $\vec{r_{o}}+\vec{l}$ , where $\ell$ is the displacement, then

$\displaystyle{ \vec{E} =\frac{1}{4 \pi \epsilon_{o}} \sum ^{n} _{i}{\frac{q_{i}(\vec{r_{o}}+\vec{l}-\vec{r_{i}})}{|\vec{r_{o}} +\vec{l} -\vec{r_{i}}|^{3}}} }$

Now if I continue according to the explanation given the textbook, then, I can take E as,

$\displaystyle{ \vec{E} = - k \vec{l} }$

where $\vec{E}$ is the field that causes a restoring force on a charge $q_{o}$ if placed at $\vec{r_{o}}+\vec{l}$ and $k$ is the a constant with dimensions $N C^{-1}m^{-1}$ and is independent of charge on the test charge, here I consider what happens if if the equilibrium was stable (doing as according to the textbook).

But I can't continue much further because I am unable to consider how to define a Gaussian surface which will not include the charges that causes the electrostatic field, also the equation that I have setup so far does not look like to help in moving towards the proof, any ideas suggestions on how to approach this problem?

  • $\begingroup$ $\displaystyle{ \vec{E} \neq - k \vec{l} }$ but $\displaystyle{ \vec{E} = \frac{- k \vec{l} }{q_{0}}}$ $\endgroup$ – Shreyansh Pathak Nov 9 '19 at 13:17

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