Question

Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert a force on each other. Half of one of the charges is transferred to the other charge and the distance between them is unchanged. What happens to the force exerted on one charge by the other charge?

Answer 1

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

$F = k \frac{q\cdot q}{r^2}$

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

$q+\frac{q}{2}=\frac{3}{2}q$

while the other charge will be

$q-\frac{q}{2}=\frac{q}{2}$

So, the new force will be

$F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F$

So, the force will decrease to 3/4 of its original value.