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Let the original charge on sphere A be q and that on B be q.

At a distance r between their centres, the magnitude of the electrostatic force on each is given by

$F=\large\frac{1}{4 \pi \in_0} \frac{qq'}{r^2}$

The sizes of spheres A and B in comparison to r are neglected. When an identical but uncharged sphere C touches A, the charges redistribute on A and C and, by symmetry, each sphere carries a charge q/2.

Similarly, after D touches B, the redistributed charge on each is q/2. Now, if the separation between A and B is halved, the magnitude of the electrostatic force on each is

$F'= \large\frac{1}{4 \pi \in_0} \frac{(q/2)(q'/2)}{(r/2)^2}$

$\quad= \large\frac{1}{4 \pi \in_0 } \frac{(qq')}{r^2}$$=F$

Thus the electrostatic force on A, due to B, remains unaltered

Hence B is the correct answer.

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