Charge and Sphere (Method of Images)

If the charge $q$ from Griffiths 3.9 is at distance $a$ from the sphere center, what sphere potential $V_0$ results in a zull force on $q$? What is the value of the charge $q$ to produce this result?

First suppose one has a grounded sphere and a point charge at distance a from the sphere center. The value of the image charge $q’$ that replaces the sphere and maintains the potential $V=0$ on all the points on the sphere surface is

$q’=-(R/a) q$ at distance $b=R^2/a$  to the right of the sphere centerYou want to add a second image charge to this setup to make the potential on the sphere $V_0 >0$. From electrostatics we know a charge exactly in the center of the sphere will generate spherical equipotential surfaces. So that we add a second image charge situated at x=0 having the value

$V_0=kq”/R$   or $q”=(RV_0)/k$

Both charges $q’$ and $q’’$ are inside the sphere, to keep it at constant potential $V_0>0$, so that for the sphere to be neutral it is necessary

$q”=-q’=(R/a) q$

The initial attraction force between $q$ and $q’$ is:

$F_1=(kqq’)/(a-b)^2 =-(kq^2 Ra)/(a^2-R^2 )^2$

After we add the second image charge the force between charge q and neutral sphere is (q and q’’ have same sign therefore the force between them is repulsive)

$F_2=F_1-(kqq”)/a^2 =-(kq^2 Ra)/(a^2-R^2)^2 +(kq^2 R)/a^3$

b)

Zero net force means:

$F_2=0$ or $a/(a^2-R^2 )^2 =1/a^3$    or $a^4/(a^2-R^2 )^2 =1$  or $a^2/(a^2-R^2 )=+/-1$
$a=R/sqrt{2}$  ,remeber this is just for finding $V_0$  here,because always $a>R$

$q”=(R/a) q=qsqrt{2}$    and $V_0=(kq”)/R=(kqsqrt{2})/R$