Question

Point charges of \(50 \mathrm{nC}\) each are located at \(A(1,0,0), B(-1,0,0), C(0,1,0)\), and \(D(0,-1,0)\) in free space. Find the total force on the charge at \(A\).

Short answer

Answer: The total force acting on the point charge at A is 5.62 x 10^-8 N, and it acts along the x-axis.

Step-by-step solution

Calculate the distance between charges

We will find the distance between point A and other points B, C, and D using the distance formula for 3D coordinates: \(r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\) Distance between A and B, \(r_{AB} = \sqrt{((-1-1)^2 + (0-0)^2 + (0-0)^2)} = 2\) Distance between A and C, \(r_{AC} = \sqrt{((0-1)^2 + (1-0)^2 + (0-0)^2)} = \sqrt{2}\) Distance between A and D, \(r_{AD} = \sqrt{((0-1)^2 + (-1-0)^2 + (0-0)^2)} = \sqrt{2}\)

Calculate individual forces acting on A

Now using Coulomb's law formula, we will calculate the forces. Force due to B on A, \(F_{AB} = k \frac{q_A q_B}{r_{AB}^2}\) \(F_{AB} = (8.99 \times 10^9) \frac{(50 \times 10^{-9})(50 \times 10^{-9})}{(2)^2} = 5.62 \times 10^{-8} N\) Force due to C on A, \(F_{AC} = k \frac{q_A q_C}{r_{AC}^2}\) \(F_{AC} = (8.99 \times 10^9) \frac{(50 \times 10^{-9})(50 \times 10^{-9})}{(\sqrt{2})^2} = 1.124 \times 10^{-7} N\) Force due to D on A, \(F_{AD} = k \frac{q_A q_D}{r_{AD}^2}\) \(F_{AD} = (8.99 \times 10^9) \frac{(50 \times 10^{-9})(50 \times 10^{-9})}{(\sqrt{2})^2} = 1.124 \times 10^{-7} N\)

Calculate the total force on A

To find the total force on A, we need to add the forces acting on A due to B, C, and D. Since the forces due to C and D are equal and along the y-axis, their combined effect along the y-axis will cancel each other out. Thus, the total force on A will be along the x-axis and equal to the force exerted by B. Total Force on A = \(F_{AB}\) Total Force on A = \(5.62 \times 10^{-8} N\) So, the total force on charge A is \(5.62 \times 10^{-8} N\) along the x-axis.