Where am I wrong in thinking this way?

The error is that it's not stable. Stable equilibrium means positive eigenvalues of the system's Hessian. Alternately phrased, stable means that any small push of the system out of equilibrium will eventually relax, and the system will be in its original state. However, in the example you gave, a small push of the positive charge will result in the charge uncontrollably accelerating towards the metal surface.

So, it's not stable.

###Specifics### Inside a conducting sphere of radius $R$ with a charge located at $\mathbf{p}$, the [potential at a point $\mathbf{r}$ induced by the surface charge will be][1] $$V(\mathbf{r})=-\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{\frac{|\mathbf{r}|^2|\mathbf{p}|^2}{R^2}+R^2-2\mathbf{r}\cdot\mathbf{p}}}$$ when $|\mathbf{r}|<R$.

Letting $\mathbf{p}=\mathbf{r}$ and noting that energy is $qV(\mathbf{r})$, we obtain $$qV(\mathbf{r})=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{\sqrt{\frac{|\mathbf{r}|^4}{R^2}+R^2-2|\mathbf{r}|^2}}=-\frac{q^2}{4 \pi R \epsilon _0}-\frac{q^2 |\mathbf{r}|^2}{4 \pi R^3 \epsilon _0}+O\left(|\mathbf{r}|^4\right)$$ which near $|\mathbf{r}|=0$ gives a harmonic oscillator with imaginary frequency (ie, exponential growth).

Hence it is unstable. [1]: http://en.wikipedia.org/wiki/Method_of_image_charges#Point_charges_2

Where am I wrong in thinking this way?

The error is that it's not stable. Stable equilibrium means positive eigenvalues of the system's Hessian. Alternately phrased, stable means that any small push of the system out of equilibrium will eventually relax, and the system will be in its original state. However, in the example you gave, a small push of the positive charge will result in the charge uncontrollably accelerating towards the metal surface.

So, it's not stable.

###Specifics### Inside a conducting sphere of radius $R$ with a charge located at $\mathbf{p}$, the potential at a point $\mathbf{r}$ induced by the surface charge will be $$V(\mathbf{r})=-\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{\frac{|\mathbf{r}|^2|\mathbf{p}|^2}{R^2}+R^2-2\mathbf{r}\cdot\mathbf{p}}}$$ when $|\mathbf{r}|<R$.

Letting $\mathbf{p}=\mathbf{r}$ and noting that energy is $qV(\mathbf{r})$, we obtain $$U(\mathbf{r})=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{\sqrt{\frac{|\mathbf{r}|^4}{R^2}+R^2-2|\mathbf{r}|^2}}=-\frac{q^2}{4 \pi R \epsilon _0}-\frac{q^2 |\mathbf{r}|^2}{4 \pi R^3 \epsilon _0}+O\left(|\mathbf{r}|^4\right)$$ which near $|\mathbf{r}|=0$ gives a harmonic oscillator with imaginary frequency (ie, exponential growth).

Hence it is unstable.

Here's what the potential surface looks like (dark is high-energy, light is low-energy):

$HistoryLength = 0;
hue = Compile[{{z, _Complex}}, {(1.0 Arg[-z] + \[Pi])/(2 \[Pi]), 
    Exp[1 - Max[Abs[z], 1]], Min[Abs[z], 1]}, 
   CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
CCompileR2[expr_] := 
  Compile[{{x, _Real}, {y, _Real}}, Evaluate[expr], 
   CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
ComplexPlotR2[f_, {x0_, x1_, \[Delta]x_}, {y0_, y1_, \[Delta]y_}, 
   mag_] := 
  Image[Transpose[
    hue[(mag f[#1[[All, All, 1]], #1[[All, All, 2]]] &)[
      Outer[List, Range[x0, x1, \[Delta]x], 
       Range[y1, y0, -\[Delta]y]]]]], ColorSpace -> Hue, 
   Magnification -> 1];
p = {x, y};
R = 1;
q = 1;
\[Epsilon]0 = 1;
V[r_] := -((
   q Boole[r.r < 1])/((4 \[Pi] \[Epsilon]0) Sqrt[(r.r p.p)/R^2 + 
     R^2 - 2 r.p]));
f = CCompileR2[q V[p] - q V[{0, 0}]];
\[Delta] = 0.00321;
L = 1.1;
ComplexPlotR2[f, {-L, L, \[Delta]}, {-L, L, \[Delta]}, 5]

Where am I wrong in thinking this way?

The error is that it's not stable. Stable equilibrium means positive eigenvalues of the system's Hessian. Alternately phrased, stable means that any small push of the system out of equilibrium will eventually relax, and the system will be in its original state. However, in the example you gave, a small push of the positive charge will result in the charge uncontrollably accelerating towards the metal surface.

So, it's not stable.

Where am I wrong in thinking this way?

The error is that it's not stable. Stable equilibrium means positive eigenvalues of the system's Hessian. Alternately phrased, stable means that any small push of the system out of equilibrium will eventually relax, and the system will be in its original state. However, in the example you gave, a small push of the positive charge will result in the charge uncontrollably accelerating towards the metal surface.

So, it's not stable.

###Specifics### Inside a conducting sphere of radius $R$ with a charge located at $\mathbf{p}$, the [potential at a point $\mathbf{r}$ induced by the surface charge will be][1] $$V(\mathbf{r})=-\frac{1}{4\pi\epsilon_0}\frac{q}{\sqrt{\frac{|\mathbf{r}|^2|\mathbf{p}|^2}{R^2}+R^2-2\mathbf{r}\cdot\mathbf{p}}}$$ when $|\mathbf{r}|<R$.

Letting $\mathbf{p}=\mathbf{r}$ and noting that energy is $qV(\mathbf{r})$, we obtain $$qV(\mathbf{r})=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{\sqrt{\frac{|\mathbf{r}|^4}{R^2}+R^2-2|\mathbf{r}|^2}}=-\frac{q^2}{4 \pi R \epsilon _0}-\frac{q^2 |\mathbf{r}|^2}{4 \pi R^3 \epsilon _0}+O\left(|\mathbf{r}|^4\right)$$ which near $|\mathbf{r}|=0$ gives a harmonic oscillator with imaginary frequency (ie, exponential growth).

Hence it is unstable. [1]: http://en.wikipedia.org/wiki/Method_of_image_charges#Point_charges_2