Question

Three point charges are positioned on the \(x\) -axis: \(+64.0 \mu \mathrm{C}\) at \(x=0.00 \mathrm{~cm},+80.0 \mu \mathrm{C}\) at \(x=25.0 \mathrm{~cm},\) and \(-160.0 \mu \mathrm{C}\) at \(x=50.0 \mathrm{~cm} .\) What is the magnitude of the electrostatic force acting on the \(+64.0-\mu C\) charge?

Short answer

Answer: The magnitude of the electrostatic force acting on the +64.0 µC charge is 5.4912 × 10^3 N.

Step-by-step solution

Identify the given information

The charges are positioned as follows: 1) Charge \(Q_1 = +64.0 \mu \mathrm{C}\) at \(x_1 = 0.00 \mathrm{~cm}\) 2) Charge \(Q_2 = +80.0 \mu \mathrm{C}\) at \(x_2 = 25.0 \mathrm{~cm}\) 3) Charge \(Q_3 = -160.0 \mu \mathrm{C}\) at \(x_3 = 50.0 \mathrm{~cm}\) We want to find the magnitude of the electrostatic force acting on charge \(Q_1\).

Calculate the distance between charges

We first need to find the distances between charge \(Q_1\) and each of the other charges. Distance between \(Q_1\) and \(Q_2\): \(d_{12} = x_2 - x_1 = 25.0 \mathrm{~cm} = 0.25 \mathrm{~m}\) Distance between \(Q_1\) and \(Q_3\): \(d_{13} = x_3 - x_1 = 50.0 \mathrm{~cm} = 0.5 \mathrm{~m}\)

Apply Coulomb's Law to find the individual forces

Coulomb's Law Formula: \(F = k\frac{|q_1q_2|}{r^2}\) where k is Coulomb's constant (\(k = 8.99×10^9 Nm^2/C^2\)), \(q_{1}\) and \(q_{2}\) are the two charges, and \(r\) is the distance between the charges. Force between \(Q_1\) and \(Q_2\): \(F_{12} = k\frac{|Q_1Q_2|}{d_{12}^2} = (8.99\times10^{9})(\frac{(64.0\times10^{-6})(80.0\times10^{-6})}{(0.25)^2})\) \(F_{12} = 1.8304 \times 10^3 \mathrm{N}\) Force between \(Q_1\) and \(Q_3\): \(F_{13} = k\frac{|Q_1Q_3|}{d_{13}^2} = (8.99\times10^{9})(\frac{(64.0\times10^{-6})(-160.0\times10^{-6})}{(0.5)^2})\) \(F_{13} = -7.3216 \times 10^3 \mathrm{N}\)

Determine the total electrostatic force

The forces \(F_{12}\) and \(F_{13}\) are acting along the x-axis, and we have their magnitudes and directions (positive/negative sign). Now we'll find the total electrostatic force acting on charge \(Q_1\) by adding the individual forces. Total Electrostatic Force: \(F_{total} = F_{12} + F_{13} = 1.8304 \times 10^3 \mathrm{N} - 7.3216 \times 10^3 \mathrm{N} = -5.4912 \times 10^3 \mathrm{N}\)

Calculate the magnitude of the total electrostatic force

The total electrostatic force on the +64.0 µC charge is \(-5.4912 \times 10^3 \mathrm{N}\). Since we only need the magnitude of this force, we can drop the negative sign, which indicates direction. Magnitude of Total Electrostatic Force: \(|F_{total}| = 5.4912 \times 10^3 \mathrm{N}\) The magnitude of the electrostatic force acting on the +64.0 µC charge is \(5.4912 \times 10^3 \mathrm{N}\).