# Tricky math step in Method of Images for conducting sphere

I am trying to understand this classic method of images problem. I have four textbooks open and they are all giving me the "it can easily be shown" for a critical math step in the solution.

For a grounded conducting sphere of radius a, and a point charge located a distance y from the center of the sphere, find the location and charge of an image charge that satisfies the boundary conditions.

After applying the boundary condition $$\phi(\vec{r}=a\hat{r}) = 0$$ you get

$$\frac{q}{|a\hat{r} - y\hat{r'}|} = \frac{-q_o}{|a\hat{r} - y'\hat{r'}|}$$

Then you factor $$a$$ from the LHS denominator and $$y'$$ from the RHS denominator. I don't know how I am supposed to see that this necessary, and most books say that this is the only factorization that "makes sense."

But also, after you do this "To obtain equality, the numerators and denominators may be set equal, respectively to obtain: $$\frac{q_o}{a} = \frac{-q'}{y'}$$" as well as another equation which can be solved simultaneously with this one to get $$y'$$.

Can someone help me see why this is apperently so obvious that \$1000 of textbooks from intro to Jacksons glance over it? Any help would be seriously appreciated. I know its probably something really trivial, but it's got me stumped bad.

Thanks

Let me rewrite your expression in a slightly different way:

$$\frac{q}{|a\hat{x} - y\hat{y}|} = \frac{-q_o}{|a\hat{x} - y'\hat{y}|}$$

So what you have to do is solve this equation:

$$|q_0\,a\hat{x} - q_0 \, y\hat{y}| = |q\,a\hat{x} - q\,y'\hat{y}|$$

Here are two vector diagrams showing the vectors $$q_0\,a\hat{x} - q_0 \, y\hat{y}$$ and $$q\,a\hat{x} - q\,y'\hat{y}$$

What one has to do is make $$BC= B'C'$$ and to do that I will make triangles $$ABC$$ and $$AB'C'$$ congruent by making $$AB = AB' \Rightarrow q_0 a = qy'$$ and $$AC = AC' \Rightarrow q_0y = aq \Rightarrow yy'=a^2$$ which gives the two equations bar a sign.

In this case y>a then if L.H.S. denominator is less than 1(one) then R.H.S. denominator must be less than 1(one) then they do so. You can take another term but this will noticed that if R.H.S. is greater than 1(one) then L.H.S must be greater than 1(one). And in the second case there is rule if (a/b)=(c/d) then we can write a=c and b=d. They do the same thing.