Let's denote the two similar conducting spheres as sphere A and sphere B. Sphere A carries charge +Q and sphere B carries charge -Q. The force of attraction between two charges can be given by Coulomb's law:
\[ F = k \frac{|q_1 q_2|}{d^2} \]
where \(F\) is the force between the charges, \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(d\) is the distance between the charges.
Initially, we have:
- Sphere A: \( +Q \)
- Sphere B: \( -Q \)
The initial attractive force is:
\[ F = k \frac{Q^2}{d^2} \]
Now, we transfer \( \frac{2}{3} \) of the charge from sphere B (which is -Q) to sphere A. The total charge transferred is:
\[ \frac{2}{3} \cdot (-Q) = -\frac{2Q}{3} \]
After the transfer, the charges on the spheres become:
- Sphere A: \( +Q - \frac{2Q}{3} = +\frac{Q}{3} \)
- Sphere B: \( -Q + \frac{2Q}{3} = -\frac{Q}{3} \)
Now, we calculate the new attractive force \( F' \) between the spheres with the new charges:
\[ F' = k \frac{\left| \frac{Q}{3} \cdot \left( -\frac{Q}{3} \right) \right|}{d^2} \]
Calculating the product of the charges:
\[ \left|\frac{Q}{3} \cdot -\frac{Q}{3}\right| = \frac{Q^2}{9} \]
Thus, the new force \( F' \) is given by:
\[ F' = k \frac{\frac{Q^2}{9}}{d^2} = \frac{1}{9} \left( k \frac{Q^2}{d^2} \right) = \frac{F}{9} \]
Therefore, the magnitude of the new attractive force is:
\[ \boxed{\frac{F}{9}} \]
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