Lately, I'm studying the method of images especially to mention the renowned example of "A point charge near a conducting sphere". So, we have a grounded conducting sphere & a point charge $\displaystyle{q}$ . The image charge, after computation, is $\displaystyle{q'} = -\frac{a}{b}\displaystyle{q}$ . Then easily we can find the field outside by using superposition principle & also find the force on $\displaystyle{q}$ solely by the field of $\displaystyle{q'}$. All is well; but then when I integrate over the whole surface, I do get the total charge on the surface of the conducting sphere is $\displaystyle{q'}$. As marked by Feynman
[...]the total induced charge is $\displaystyle{q'}$, as it should be.
I didn't get at the physical reasoning behind getting the same magnitude of charge as that of image charge. The image charge was located at $\dfrac{a^2}{b}$ w.r.t. the center of the sphere while the surface is located at $a$. So, the distance between image charge & $q$ is $\left(b - \dfrac{a^2}{b}\right)$, while that of the surface & $q$ is $(b - a)$; so though the magnitude of the charge on the surface is same as that of the image charge, the distance is not the same. So, how can the fields be same? In order to have the zero potential, charge of $\displaystyle{q'}$ must be located at a distance of $\dfrac{a^2}{b}$ from the center of the sphere. If $q'$ be situated at some other position like at the surface which is not at $\dfrac{a^2}{b}$ but at $a$, then how can there be same field, after all the distance has been changed between $q$ & $q'$? I'm not understanding, thus why the charge at the surface is $\displaystyle{q'}$. Can anyone please explain me the reason??