Shown are three separate pairs of point charges. Assume the pairs interact only with each other. Rank the magnitudes of the force between the pairs from largest to smallest.

Solution 2R STEP 1: In the 1st pair the charges are, Q = 4 q , Q = + 2 q 1 2 The distance between them is “x”. The force between them according to coulomb’s law will be, Q Q F = 1 122 40 r 1 (4q)×2q 9 (8q ) = 40 x2 = 9 × 10 × x2 9 2 2 = 72 × 10 × q / x The minus sign indicates the force is attractive. 9 2 2 The magnitude is 72 × 10 × q / x . STEP 2: For the 2nd pair of charges, Q 1 + 3 q , Q = + 32q The distance between them is “x”. The force between them according to coulomb’s law will be, F = 1 Q 122 40 r 1 3q×3q 9 9q2 = 4 x2 = 9 × 10 × x2 0 = 81 × 10 × q / x 2 2 The magnitude is 81 × 10 × q / x . 2 2 STEP 3:- For the 2nd pair of charges, Q 1 2 q , Q = 2 q The distance between them is “x/2”. The force between them according to coulomb’s law will be, 1 Q 1 2 F = 4 r2 0 2 = 1 2q×(2q) = 9 × 10 ×9 4 q 40 (x/2) (x/2) = 36 × 10 × 4 × q / x 2 2 9 2 2 = 144 × 10 × q / x 9 2 2 The magnitude is 144 × 10 × q / x . CONCLUSION: So the the force between the 3rd pair of charges is maximum. Then the force between 2nd pair of charges will come. The least force is between the 1st pair of charges.