7.1 State Coulomb's law in words - NSC Physical Sciences - Question 7 - 2017 - Paper 1

The sign of charge Q2 is negative. This is concluded because the net force on charge Q3 is directed to the left, indicating that Q2 creates an attractive force.

To find the magnitudes of Q1 and Q2, we use Coulomb's law in the context of the forces acting on Q3:

1. The force experienced by Q3 due to Q1 is given by: $F = k \frac{Q_1 Q_3}{d^2}$ Substituting the values: $0,012 = (9 \times 10^9) (Q_1)(2 \times 10^{-6}) / (2,5)^2$ Rearranging gives: $Q_1 = \frac{0.012 \times (2.5)^2}{9 \times 10^9 \times 2 \times 10^{-6}} = 4,17 \times 10^{-6} C$

2. Next, we calculate Q2 using the previously found equation: $0.3 = k \frac{Q_2 (2 \times 10^{-6})}{1^2}$ Rearranging gives: $Q_2 = \frac{0.3 \times 1^2}{9 \times 10^9 \times 2 \times 10^{-6}} = 1,73 \times 10^{-5} C$