Question

Two identical metal block with charges +2Q and -Q are separated by some distance, and exert a force F on each other. The force between them then will be.

Answer 1

Given:
One block has a charge of +2Q and the other block has a charge of -Q 
Using Coulomb's law, the force between two charges is given by: \(F = \frac{k|q_1q_2|}{r^2}\)
Where: \( F \) = force between the charges 
K = Coulomb's constant (approximately \(( 8.9875 \times 10^9 \, \text{N.m}^2/\text{C}^2 )) \)
\(q_1\) and \(q_2\)= the two charges 
\( r \) = distance between the charges 
For our problem: \(( q_1 = +2Q ) and ( q_2 = -Q )\)
The force between them
\(F' = \frac{k(2Q)(-Q)}{r^2}\)
\(F' = \frac{-2kQ^2}{r^2}\)
Given that they initially exert a force F on each other:
\(F = \frac{k|q_1q_2|}{r^2} = \frac{2kQ^2}{r^2}\)
Comparing the two forces:
\(F' = -F\)
The negative sign indicates that the direction of the force is reversed. But since the magnitude of the force remains the same, the force between them will still be  F , but in the opposite direction