Self-energy of conducting shell While calculating self energy of a conducting shell we integrate $Vdq$,where $V=\frac{q}{4π\epsilon r}$, but when the $dq$ charge is bought close to the shell it changes the charge distribution on the sphere and hence does work on the charge already present on the sphere..... moreover due to the changing charge distribution, the expression of potential changes too.....I agree that the effects above are negligible for a small dq charge but how do we conclude that their contribution is still negligible after we add up the effects for all the $dq$'s (I mean integrating over all the charges).
 A: Assume spherical conducting shell with radius $r_0$ and charge $q$. Let's see the minimum work that needs to be done to bring a infinitesimal charge $dq$ with initial kinetic energy $0$.
Let's look at the moment at which $dq$ is separated from the center of the shell by distance $r$. The charge distribution on the shell can be represented by 2 point charges of following amounts (from here):
One as the image of the dq at the following distance inside the sphere, and the other on the center to balance the total charge on the shell to $q$:
$$dq' = -dq\,\frac{r_0}{r},\ \ r_{dq'} = \frac{{r_0}^2}{r},\ \ q_{center} = q + dq\,\frac{r_0}{r}$$
Writing the potential energy at this state:
$$dU = \frac{1}{4\pi\epsilon_0}\,\left(
\frac{-dq\,\frac{r_0}{r}\cdot dq}{(r-\frac{{r_0}^2}{r})} +
\frac{-dq\,\frac{r_0}{r}\cdot (q+dq\,\frac{r_0}{r})}{\frac{{r_0}^2}{r}} +
\frac{dq\cdot (q+dq\,\frac{r_0}{r})}{r} \right)$$
Then, if we maximize this energy with respect to $r$ on region $(r_0,\infty)$, we find the minimum energy needed to make this $dq$ to go join $q$.
First, rewrite $dU$ for simplicity (assuming the second degree differentials are irrelevant ($dq^2$)),
$$dU = \frac{dq}{4\pi\epsilon_0}\,\left( 0 +
\frac{-q\,\frac{r_0}{r}}{(\frac{{r_0}^2}{r})} +
\frac{q}{r} \right)$$
$$dU = \frac{dq}{4\pi\epsilon_0}\,\left( -\frac{q}{r_0} +
\frac{q}{r} \right)$$
Since this is a monotonically decreasing function of $r$, we can say its maximum value in the specified region is when $r=r_0$.
This is a counter-intuitive result, since
$$\lim_{r\to\infty} dU = -\frac{dq}{4\pi\epsilon_0}\frac{q}{r_0}$$
and
$$\lim_{r\to r_0^+} dU = 0$$
Since these are the initial and maximum potentials of the trajectory, the work done needs to be
$$dW = \left(\lim_{r\to r_0^+} dU\right) - \left(\lim_{r\to\infty} dU\right) = \frac{dq}{4\pi\epsilon_0}\frac{q}{r_0}$$
But this does show us this: even though our calculations were different and not intuitive, the final result is the same.
