Question

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?

Answer 1

Answer is it will decrease to one-sixth the original force. or C.

Answer 2

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.