One charge is decreased to one-third of its original value, and a second charge is decreased to one-half of its original value. How will the electrical force between the charges compare with the original force?

The original Coulomb force between the charges is:

Fc = (k*Q₁*Q₂) / r², where k is the Coulomb constant and k=9*10⁹ N m² C⁻², Q₁ is the first charge, Q₂ is the second charge and r is the distance between the charges.

The magnitude of the force is independent of the sign of the charge so I can simply say they are both positive.

Q₁ is decreased to Q₁₁ = (1/3) * Q₁=Q₁/3 and

Q₂ is decreased to Q₂₂ = (1/2) * Q₂=Q₂/2.

New force:

Fc₁ = (k*Q₁₁*Q₂₂) r², now we input the decreased values of the charge

Fc₁ = (k*{Q₁/3}*{Q₂/2}) / r², that is equal to:

Fc₁ = (k * (1/3) * (1/2) * Q₁*Q₂) / r²,

Fc₁ = (k * (1/6) * Q₁*Q₂) / r²

Fc₁ = (1/6) * (k*Q₁*Q₂) / r², and since the original force is: Fc = (k*Q₁*Q₂) / r² we get:

Fc₁ = (1/6) * Fc

So the magnitude of the new force Fc₁ with decreased charges is 6 times smaller than the original force Fc.