Question

Two like charges balloons placed at a distance of 0.50 m, experience a repulsive force of 0.32 n. what is the force if the distance between the balloons is doubled

Answer 1

The electrostatic force between two charged object is given by:

`F=k (q_1 q_2)/(r^2)`
where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the separation between the two objects

We see that the force is inversely proportional to the square of the distance:
`(1)/(r^2)`. Therefore, if the distance is doubled, the force decreases by a factor 4, and the new force will be:

`F'= (F)/(4)= (0.32 N)/(4)=0.08 N`
and it will still be a repulsive force, since the two balloons have charges of same sign.