Answer:
The new force is 1/4 of the previous force.
Explanation:
Given
[tex]Initial\ Distance = 2cm[/tex] ---- [tex]r_1[/tex]
[tex]New\ Distance = 4cm[/tex] --- [tex]r_2[/tex]
Required
Determine the new force
Let the two particles be q1 and q2.
The initial force F1 is:
[tex]F_1 = \frac{kq_1q_2}{r_1^2}[/tex] --- Coulomb's law
Substitute 2 for r1
[tex]F_1 = \frac{kq_1q_2}{2^2}[/tex]
[tex]F_1 = \frac{kq_1q_2}{4}[/tex]
The new force (F2) is
[tex]F_2 = \frac{kq_1q_2}{r_2^2}[/tex]
Substitute 4 for r2
[tex]F_2 = \frac{kq_1q_2}{4^2}[/tex]
[tex]F_2 = \frac{kq_1q_2}{4*4}[/tex]
[tex]F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}[/tex]
Substitute [tex]F_1 = \frac{kq_1q_2}{4}[/tex]
[tex]F_2 = \frac{1}{4}*F_1[/tex]
[tex]F_2 = \frac{F_1}{4}[/tex]
The new force is 1/4 of the previous force.
