Ans: E

When sphere III, which is initially uncharged, is touched to sphere I, it will share the charge equally with sphere I. Similarly, when sphere III is touched to sphere II, the charges will redistribute. We need to find the final charges on spheres I and II and then determine the new electrostatic force between them.

1. Initial Condition:
Spheres I and II each have a charge \(+Q\).
Sphere III has a charge of 0.

2. First Touch:
Sphere III is touched to sphere I.
The total charge between sphere I and sphere III is \(Q + 0 = Q\).
This charge is shared equally because the spheres are identical, so each will have a charge of \(\frac{Q}{2}\).

After the first touch:
Sphere I: \(\frac{Q}{2}\)
Sphere III: \(\frac{Q}{2}\)
Sphere II: \(Q\)

3. Second Touch:
 Sphere III (now with charge \(\frac{Q}{2}\)) is touched to sphere II (with charge \(Q\)).
 The total charge between sphere II and sphere III is \(Q + \frac{Q}{2} = \frac{3Q}{2}\).
 This charge is shared equally, so each will have a charge of \(\frac{3Q}{4}\).

After the second touch:
 Sphere I: \(\frac{Q}{2}\) (unchanged from the first touch)
 Sphere II: \(\frac{3Q}{4}\)
Sphere III: \(\frac{3Q}{4}\)

New Force Calculation:
 The new charges on sphere I and sphere II are \(\frac{Q}{2}\) and \(\frac{3Q}{4}\), respectively.

The electrostatic force between two charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by Coulomb’s law:

\[
F = k_e \frac{q_1 q_2}{r^2}
\]

Initially, the force \(F\) is given by:

\[
F = k_e \frac{Q \cdot Q}{r^2} = k_e \frac{Q^2}{r^2}
\]

With the new charges, the force \(F’\) is:

\[
F’ = k_e \frac{\left(\frac{Q}{2}\right) \left(\frac{3Q}{4}\right)}{r^2} = k_e \frac{\frac{3Q^2}{8}}{r^2} = \frac{3}{8} \left( k_e \frac{Q^2}{r^2} \right) = \frac{3}{8} F
\]

The new magnitude of the electrostatic force between spheres I and II is: \[ \frac{3}{8} F \]