Why is the force on an image charge the same as the force on the object it replaces?

Consider the arrangement of a point charge $$q$$ situated a distance $$a$$ from the center of a grounded conducting sphere of radius R (Griffith 4 th Ed. Example 2 or here). We find that we can replace the sphere by a image charge $$q' = -Rq/a$$, placed a distance $$b = R^2/a$$ from the center of the sphere.

However, how do we calculate the force on the sphere? Griffith states that it is equivalent to the force on the image charge $$q'$$ but what is the justification for this?

• Using image theory you’re always solving a boundary value problem equivalent to the original problem. In other words the original problem and image problem should be the same in the region where the image charge doesn’t exist. It may be a good idea to look online to find some material on the background mathematical theory of the image method. Jan 13 at 11:49
• I missed this explanation: in this problem the boundary is the sphere and the 2 regions are outside and inside of the sphere. You want to design an equivalent problem outside of the sphere using image theory. The laws of electromagnetics are the same on both side of the sphere in both problems. Thus, once you come up with a configuration that causes the boundary value (of potential) to be the same (on the sphere) then due to uniqueness, you’ll get the result in the image problem which is equivalent to the solution of the original problem. Jan 13 at 12:03

• By definition, the image charge produces the same field $$\mathbf{E}_1$$ outside the sphere as the real arrangement of surface charge does. That's the entire point of image charges.
• Since the force on the real charge is $$q \mathbf{E}_1$$, the force of the image charge on the real charge is the same as the force of the actual surface charge on the real charge.