Four point charges are placed at the following \(x y\) coordinates: \(Q_{1}=-1 \mathrm{mC},\) at \((-3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{2}=-1 \mathrm{mC},\) at \((+3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{3}=+1.024 \mathrm{mC},\) at \((0 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{4}=+2 \mathrm{mC},\) at \((0 \mathrm{~cm},-4 \mathrm{~cm})\) Calculate the net force on charge \(Q_{4}\) due to charges \(Q_{1}, Q_{2}\) and \(Q_{3}\).

First, we need to convert all given quantities, charges and coordinates to SI units. -1 mC = -1 × 10⁻³ C 1.024 mC = 1.024 × 10⁻³ C 2 mC = 2 × 10⁻³ C All the coordinates are given in centimeters. We need to convert them to meters. -3 cm = -0.03 m +3 cm = 0.03 m 0 cm = 0 m -4 cm = -0.04 m

We need to calculate the distance between \(Q_{4}\) and each of the other charges using the distance formula. $$ d_{ij} = \sqrt{(x_{i}-x_{j})^{2}+(y_{i}-y_{j})^{2}} $$ For \(Q_{1}\) and \(Q_{4}\): $$ d_{14} = \sqrt{(-0.03 - 0)^{2} + (0 - (-0.04))^{2}} = 0.05 \textrm{ m} $$ For \(Q_{2}\) and \(Q_{4}\): $$ d_{24} = \sqrt{(0.03 - 0)^{2} + (0 - (-0.04))^{2}} = 0.05 \textrm{ m} $$ For \(Q_{3}\) and \(Q_{4}\): $$ d_{34} = \sqrt{(0 - 0)^{2} + (0 - (-0.04))^{2}} = 0.04 \textrm{ m} $$

Using Coulomb's Law, we'll calculate the force between each pair of charges. $$ F_{ij} = k\frac{Q_{i}Q_{j}}{d_{ij}^{2}} $$ Where \(k = 8.9875517923 × 10^{9} \textrm{ N m² C⁻²}\) For \(Q_{1}\) and \(Q_{4}\): $$ F_{14} = 8.988×10^{9} × \frac{(-1×10^{-3})(2×10^{-3})}{0.05^{2}} = -7.19 \textrm{ N} $$ For \(Q_{2}\) and \(Q_{4}\): $$ F_{24} = 8.988×10^{9} × \frac{(-1×10^{-3})(2×10^{-3})}{0.05^{2}} = -7.19 \textrm{ N} $$ For \(Q_{3}\) and \(Q_{4}\): $$ F_{34} = 8.988×10^{9} × \frac{(1.024×10^{-3})(2×10^{-3})}{0.04^{2}} = 11.6 \textrm{ N} $$

Now that we have the magnitude of the force for each pair of charges, we need to determine the directions for each force by using the angle between each charge pair. First, we need to find the angle between \(Q_{4}\) and the x-axis for both \(Q_{1}\) and \(Q_{2}\). For \(Q_{1}\) and \(Q_{4}\): $$\tan{\theta_{1}} = \frac{-0.04}{-0.03}$$ $$\theta_{1} = \arctan{\frac{-0.04}{-0.03}} \approx 53.13^{\circ}$$ For \(Q_{2}\) and \(Q_{4}\): $$\tan{\theta_{2}} = \frac{-0.04}{0.03}$$ $$\theta_{2} = \arctan{\frac{-0.04}{0.03}} \approx -53.13^{\circ}$$

Knowing the magnitude and angle of each force, we can compute the \(x\) and \(y\) components of each force. For \(F_{14}\): $$F_{14x} = F_{14} \cos{53.13^{\circ}} = 4.14 \textrm{ N}$$ $$F_{14y} = F_{14} \sin{53.13^{\circ}} = 5.52 \textrm{ N}$$ For \(F_{24}\): $$F_{24x} = F_{24} \cos{-53.13^{\circ}} = 4.14 \textrm{ N}$$ $$F_{24y} = F_{24} \sin{-53.13^{\circ}} = -5.52 \textrm{ N}$$ For \(F_{34}\): $$F_{34x} = 0 \textrm{ N}$$ $$F_{34y} = 11.6 \textrm{ N}$$

Now that we have the components for each force, we can add them together to find the net force on \(Q_{4}\). $$F_{net_x} = F_{14x} + F_{24x} + F_{34x} = 4.14 \textrm{ N} + 4.14 \textrm{ N} + 0 \textrm{ N} = 8.28 \textrm{ N}$$ $$F_{net_y} = F_{14y} + F_{24y} + F_{34y} = 5.52 \textrm{ N} - 5.52 \textrm{ N} + 11.6 \textrm{ N} = 11.6 \textrm{ N}$$ Finally, we have the net force in both components, so we can combine them to find the magnitude and direction of the net force. To find the magnitude: $$F_{net} = \sqrt{F_{net_x}^{2} + F_{net_y}^{2}} = \sqrt{(8.28 \textrm{ N})^{2} + (11.6 \textrm{ N})^{2}} = 14.3 \textrm{ N}$$ To find the direction of the net force, we can compute the angle: $$\tan{\theta_{net}} = \frac{F_{net_y}}{F_{net_x}}$$ $$\theta_{net} = \arctan{\frac{11.6}{8.28}} \approx 54.5^{\circ}$$ So the net force on charge \(Q_{4}\) is \(14.3 N\) at an angle of \(54.5^\circ\).