The diagram shows a particle with charge +Q and a particle with charge -Q separated by a distance d - AQA - A-Level Physics - Question 14 - 2021 - Paper 2

To find the new force between the two charged particles after transferring an additional charge of +2Q to each, we can use Coulomb's Law, which states:

$F = k \frac{|q_1 q_2|}{r^2}$

where:

• $F$ is the force between the charges,
• $k$ is Coulomb's constant,
• $q_1$ and $q_2$ are the magnitudes of the two charges,
• $r$ is the distance between the centers of the two charges.

1. Initial Conditions: Initially, we have charges of +Q and -Q with distance d.

   • The initial force (attraction) is given by: $F = k \frac{Q \cdot Q}{d^2} = k \frac{Q^2}{d^2} ext{ (attractive force)}$

2. New Conditions: After adding +2Q to each particle, the charges now are +3Q and +Q.

   • The new force (after charge addition) is: $F' = k \frac{(3Q)(Q)}{(2d)^2} = k \frac{3Q^2}{4d^2}$

3. Comparing Forces: Now we compare this with the initial force:

   • The original force F was: $F = k \frac{Q^2}{d^2}$
   • The new force can be expressed in terms of F: $F' = \frac{3}{4} \times F$

The force now acts between the particles is a repulsive force of ( \frac{3}{4}F ).