Question

Point charges of \(1 \mathrm{nC}\) and \(-2 \mathrm{nC}\) are located at \((0,0,0)\) and \((1,1,1)\), respectively, in free space. Determine the vector force acting on each charge.

Short answer

In summary, the vector force acting on the \(1 \mathrm{nC}\) charge is \(3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\mathrm{N}\) and the vector force acting on the \(-2 \mathrm{nC}\) charge is \(-3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\mathrm{N}\). These two forces are equal in magnitude but opposite in direction.

Step-by-step solution

Understand Coulomb's Law

Coulomb's law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it can be written as: \(F = k\frac{|q_1q_2|}{r^2}\) where \(F\) is the electrostatic force, \(k\) is Coulomb's constant (\(8.99 × 10^9 Nm^2/C^2\)), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between them.

Convert the charges to Coulombs

The charges given are in nanocoulombs (nC). We should first convert them to Coulombs (C) before proceeding: \(1 \mathrm{nC} = 10^{-9} \mathrm{C}\) So, \(q_1 = 1 \times 10^{-9} \mathrm{C}\) and \(q_2 = -2 \times 10^{-9} \mathrm{C}\).

Calculate the distance between the charges

Since the charges are located at \((0,0,0)\) and \((1,1,1)\), we can use the distance formula to find the distance between them: \(r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} = \sqrt{(1-0)^2 + (1-0)^2 + (1-0)^2} = \sqrt{3}\)

Apply Coulomb's Law to find the magnitude of force

Using the values of \(q_1\), \(q_2\), and \(r\), we can now calculate the electrostatic force between the charges: \(F = k\frac{|q_1q_2|}{r^2} = 8.99 × 10^9 \frac{|1 \times 10^{-9} \times (-2) \times 10^{-9}|}{(\sqrt{3})^2} = 3.99 \times 10^{-9} \mathrm{N}\)

Find the unit vector

To find the vector force acting on each charge, we need to determine the unit vector from \((0,0,0)\) to \((1,1,1)\). The unit vector can be calculated as: \(\hat{r} = \frac{(1-0)\hat{i} + (1-0)\hat{j} + (1-0)\hat{k}}{\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)

Determine the vector force

Finally, we can find the vector force acting on each charge by multiplying the electrostatic force magnitude \(F\) with the unit vector: \(F_{12} = F\hat{r} = 3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\) Since the force acting on the first charge is equal and opposite to the force acting on the second charge, we have: \(F_{21} = -F\hat{r} = -3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\) So the vector force acting on the \(1 \mathrm{nC}\) charge at \((0,0,0)\) is \(3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\mathrm{N}\). The vector force acting on the \(-2 \mathrm{nC}\) charge at \((1,1,1)\) is \(-3.99 \times 10^{-9} \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\mathrm{N}\).