If we place a charge +q at a distance a along the x-axis from a grounded plate, we can use the method of images to show the potential is $V(\textbf{r})=\frac{q}{4\pi \epsilon_0}(\frac{1}{r_1} -\frac{1}{r_2})$, where $\frac{1}{r_1}$ & $\frac{1}{r_2}$ are the distances from +q and the image charge -q respectively. Following this reasoning, my notes(and Wikipedia) state that from $\textbf{E}=-\bigtriangledown V$ that $\textbf{E}_{x=0} = \frac{-qa}{2 \pi \epsilon_0 (a^2+y^2+z^2)^{\frac{3}{2}}}$, so by $E=\frac{\sigma}{\epsilon_0}$ for a charged plate, the surface charge density is $\sigma=\frac{-qa}{2 \pi (a^2+y^2+z^2)^{\frac{3}{2}}}$. However, doesn’t this use the total field of the point charge and the plate, when we should be using only the field produced by the plate to calculate the surface charge density?