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