Equal Charges May Attract

Here is a brief sketch of the problem
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This problem requires the knowledge of Method of Images , which can be referred to from this link.

Using Method of Images, we can find the images of the point charge in the sphere, which results in the following situation,
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Now,
Consider the total charge on the conducting sphere in the above figure.

It is obvious that, according to the figure, the total charge on the sphere is $\displaystyle q' = -\frac{R}{d}q$ .

But, since the sphere is isolated, the total charge on it must remain same. In other words, the final charge on the sphere must be the same as $\displaystyle +q$

Hence, to make the necessary corrections, we introduce another positive charge, which is placed at the center of the sphere (to distribute the potential uniformly) and has a magnitude of $\displaystyle Q$ .
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Using the fact that the sphere is isolated,
$Q+q' = q$

Thus,
$Q = q + \frac{R}{d}q = \left(1+\frac{R}{d}\right)q$

Now, we need to find the force on $\displaystyle q$ due to $\displaystyle Q$ and $\displaystyle q'$ , which is,

$F = \frac{kQq}{d^2} - \frac{kqq'}{(d-\frac{R^2}{d})^2}$

When the charges are just about to start attracting, its clear that $\displaystyle F=0$ .

$\frac{kqQ}{d^2} = \frac{kqq'}{(d-\frac{R^2}{d})^2}$

Substituting the values of the charges, and simplifying the expression, we get,

$\left(1+\frac{d}{R}\right)\left(\frac{d^2}{R^2}-1\right)^2 = \left(\frac{d}{R}\right)^4$

Solving for $\displaystyle \frac{d}{R}$ , we find that,

$\frac{d}{R} = 1.618, 0.75488$

We neglect the second solution, because if $\displaystyle \frac{d}{R} = 0.75488$ , then, $\displaystyle R>d$

But, the problem already mentions that $\displaystyle d>R$ . This is a contradiction.

Hence,
$\frac{d}{R} = \boxed{1.618}$