The force between two charges is 10 n. if the distance between the two charges is halved, what is the new force between the charges?

16N

Explanation:

According to Coulomb's law, the force existing between two charges is directly proportional to the product of their charges and inversely proportional to the distance between the charges. Mathematically,

F = kq1q2/r²

Where F is the force between them

q1 and q2 are the charges

r is the distance between them.

If the force between them is 10N, the formula becomes;

10 = kq1q2/r²... (1)

If the distance between them is now halved, the force will become

F = kq1q2/(r/2)²

F = kq1q2/(r²/4)

F = 4kq1q2/r² ... (2)

Dividing equation 1 by 2 we have;

4/F = (kq1q2/r²)÷ (4kq1q2/r²)

4/F = kq1q2/r²×r²/4k1q1q2

4/F = 1/4

Cross multiplying we have;

F = 4×4

F = 16N

Therefore the new force between the charges is 16N

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