Force cap f acts between two charges, q sub 1 and q sub 2 , separated by a distance d If q sub 1 is increased to twice its original value and the distance between the charges is also doubled, what is the new force acting between the charges in terms of cap f ?

By Coulomb’s law, the force is proportional to the product of the charges and inversely proportional to the square of the distance. Doubling q₁ multiplies the force by 2, while doubling the distance divides it by 2² = 4. The new force is therefore (2/4)f = ½f, so the answer is “1 half f.”

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