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
The electrostatic force between two charged object is given by: [tex]F=k \frac{q_1 q_2}{r^2} [/tex] 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: [tex] \frac{1}{r^2} [/tex]. Therefore, if the distance is doubled, the force decreases by a factor 4, and the new force will be: [tex]F'= \frac{F}{4}= \frac{0.32 N}{4}=0.08 N [/tex] and it will still be a repulsive force, since the two balloons have charges of same sign.
