Question

Four point charges are placed at the following \(x y\) -coordinates: \(Q_{1}=-1.00 \mathrm{mC},\) at \((-3.00 \mathrm{~cm}, 0.00 \mathrm{~cm})\) \(Q_{2}=-1.00 \mathrm{mC}_{2}\) at \((+3.00 \mathrm{~cm}, 0.00 \mathrm{~cm})\) \(Q_{3}=+1.024 \mathrm{mC},\) at \((0.00 \mathrm{~cm}, 0.00 \mathrm{~cm})\) \(Q_{4}=+2.00 \mathrm{mC},\) at \((0.00 \mathrm{~cm},-4.00 \mathrm{~cm})\) Calculate the net force on charge \(Q_{4}\) due to charges \(Q_{1}, Q_{2},\) and \(Q_{3}\)

Short answer

Answer: The net force acting on charge \(Q_4\) can be calculated by finding the magnitudes and components of the forces due to the other charges (\(Q_1\), \(Q_2\), and \(Q_3\)), and then adding them up as vectors: \(F_{net_x} = -F_{41} \frac{0.03}{0.05} + F_{42} \frac{0.03}{0.05}\) and \(F_{net_y} = F_{41} \frac{0.04}{0.05} + F_{42} \frac{0.04}{0.05} - F_{43}\).

Step-by-step solution

Convert distances to meters

Convert the distance from cm to meters: \(Q_1\): \((-0.03\ \text{m}, 0\ \text{m})\) \(Q_2\): \((0.03\ \text{m}, 0\ \text{m})\) \(Q_3\): \((0\ \text{m}, 0\ \text{m})\) \(Q_4\): \((0\ \text{m}, -0.04\ \text{m})\)

Calculate distance between each pair of charges

Find the distances between \(Q_4\) and each of the other charges: \(d_{41} = \sqrt{(-0.03 - 0)^2 + (0 - (-0.04))^2} = 0.05\ \text{m}\) \(d_{42} = \sqrt{(0.03 - 0)^2 + (0 - (-0.04))^2} = 0.05\ \text{m}\) \(d_{43} = \sqrt{(0 - 0)^2 + (0 - (-0.04))^2} = 0.04\ \text{m}\)

Calculate force between each pair of charges using Coulomb's law

The Coulomb's law gives the force between two charges as \(F = k\frac{|q_1q_2|}{r^2}\), where \(k = 8.99 \times 10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2\) is the electrostatic constant. Find the magnitudes of the forces acting on \(Q_4\) due to the other charges: \(F_{41} = k\frac{|Q_4Q_1|}{d_{41}^2} = 8.99 \times 10^{9} \cdot \frac{2 \cdot 10^{-6} \cdot 1 \cdot 10^{-3}}{(0.05)^2}\ \text{N}\) \(F_{42} = k\frac{|Q_4Q_2|}{d_{42}^2} = 8.99 \times 10^{9} \cdot \frac{2 \cdot 10^{-6} \cdot 1 \cdot 10^{-3}}{(0.05)^2}\ \text{N}\) \(F_{43} = k\frac{|Q_4Q_3|}{d_{43}^2} = 8.99 \times 10^{9} \cdot \frac{2 \cdot 10^{-6} \cdot 1.024 \cdot 10^{-3}}{(0.04)^2}\ \text{N}\)

Calculate the x and y components of each force

Find the x and y components of the forces: \(F_{41x} = F_{41} \cos{(180^\circ - \theta)} = -F_{41} \frac{0.03}{0.05}\) \(F_{41y} = F_{41} \sin{(180^\circ - \theta)} = F_{41} \frac{0.04}{0.05}\) \(F_{42x} = F_{42} \cos{\theta} = F_{42} \frac{0.03}{0.05}\) \(F_{42y} = F_{42} \sin{\theta} = F_{42} \frac{0.04}{0.05}\) \(F_{43x} = 0\) (since \(Q_4\) and \(Q_3\) are along the same vertical line) \(F_{43y} = -F_{43}\) (since the force is in the negative y direction)

Calculate the net force acting on \(Q_4\)

Calculate the net force by adding the components of each force: \(F_{net_x} = F_{41x} + F_{42x} + F_{43x} = -F_{41} \frac{0.03}{0.05} + F_{42} \frac{0.03}{0.05}\) \(F_{net_y} = F_{41y} + F_{42y} + F_{43y} = F_{41} \frac{0.04}{0.05} + F_{42} \frac{0.04}{0.05} - F_{43}\) After calculating those forces and adding their components, we will get the net force acting on charge \(Q_4\).