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?

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 $\textbf{E}=-\nabla V$ that

$$\textbf{E}_{x=0} = \frac{-qa}{2 \pi \epsilon_0 (a^2+y^2+z^2)^{3/2}}$$

So by $E=\frac{\sigma}{\epsilon_0}$ for a charged plate, the surface charge density is $$\sigma=\frac{-qa}{2 \pi (a^2+y^2+z^2)^{3/2}}$$

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?

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 $E=\frac{q}{\epsilon_0}$ that the surface charge 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?

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?