Two charges are sitting 1.5 m apart with a force of 3 N between them. They are now moved farther apart to 2.25 m and one of the charges is increased by a factor of 4. What is the magnitude of the new force between the two charges?

Answer: 2.37N

Explanation:

According to coulombs law which states that the force of attraction (F) between two charges (q1 and q2) is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them. Mathematically,

F = kq1q2/r²

For the first two charges that are sitting 1.5 m apart with a force of 3 N between them, we have

3 = kq1q2/1.5²

3 = kq1q2/2.25

Kq1q2 = 6.75 ... (1)

If the charges are now moved farther apart 2.25 m and one of the charges is increased by a factor of 4. The formula becomes

F2 = k (4q1) q2/2.25² (q1 has been increased by factor of 4)

k (4q1) q2 = 5.06F2 ... (2)

Dividing 2 by 1 we have

k (4q1) q2/kq1q2 = 5.06F2/3

4 = 5.06F2/3

5.06F2 = 12

F2 = 12/5.06

F2 = 2.37N

Therefore the magnitude of the new force between the two charges is 2.37N