Answer:
The new electric force on object B by object A is one half of the magnitude of the original force F, in the opposite direction. Equals -F/2
Explanation:
We are asked to demonstrate what happens when you reduce the magnitude of the charge of an object in an interactive system.
Assumptions.
The new force is F'
The system is isolated from other external interferences
The distance between objects A and B remains unchanged
Analysis.
The force F on object A by object B is given by the formula
F=(k*q,a*q,b)/d^2
where k is the Coulomb's Law constant, and equals 9x10^9 N/m s
q,a and q,b equals the original charges of objects A and B
d is the distance between objects A and B
Now, we are to calculate the new force, let's call it F', and following the same reasoning.
F'=(k*q',a*q',b)/d'^2
Given that only q', b is different to the original, we have that
F'=(k*q,a*q',b)/d^2
Now, q',b=q,b/2 and replacing
F'=(k*q,a*q,b)/(2*d^2)
Now we find the relation between F' and F
F'/F=[(k*q,a*q,b)/(2*d^2)]/[(k*q,a*q,b)/d^2]
Reducing similar terms
F'/F=1/2
However, we should also consider that the forces F and F' are in oposite directions, so simplifying the vectorial notation,
F'/F=-1/2, meaning that F'=-F/2
