Question

Two point charges, with charge magnitudes q and ????, are placed a distance r apart. In this arrangement, each point charge experiences a force of magnitude F. In the three scenarios, q, ????, and r are changed to some multiple of their original values. In each case, fill in the missing value on one of the variables that will result in a force between the point charges that is equal in magnitude to the original F.
Original q Q r
Scenario 1 2q ...Q 4r
Scenario 2 ...q 2Q (1/2)r
Scenario 3 (3/2)q (3/8)Q ...r

Answer 1

Answer:

1)  Q ’= 8 Q ,  2)    q ’= 16 q ,  3)   r ’= ¾ r

Explanation:

For this exercise we will use Coulomb's law

      F = k q Q / r²

It asks us to calculate the change of any of the parameters so that the force is always F

Original values

                q, Q, r

Scenario 1

      q ’= 2q

       r ’= 4r

     F = k q ’Q’ / r’²

we substitute

     F = k 2q Q ’/ (4r)²

     F = k 2q Q '/ 16r²

we substitute the value of F

      k q Q / r² = k q Q '/ 8r²

       Q ’= 8 Q

Scenario 2

       Q ’= Q

       r ’= 4r

we substitute

      F = k q ’Q / 16r²

      k q Q / r² = k q’ Q / 16 r²

      q ’= 16 q

Scenario 3

      q ’= 3/2 q

      Q ’= ⅜ Q

we substitute

        k q Q r² = k (3/2 q) (⅜ Q) / r’²

        r’² = 9/16 r²

        r ’= ¾ r