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.

Key concepts

Point Charge

A point charge is a charge that is localized to a single point in space. It's like imagining all the electric charge concentrated in one tiny spot. This simplification helps us calculate forces and fields easily. In our scenario, we have point charges each with a magnitude of 50 nanocoulombs (nC), situated at specific coordinates. Understanding that these charges are at "points" means we can use specific mathematical formulas to calculate interactions between them.

Distance Formula

To find the force between point charges, first, we need to know the distance between them. This is where the distance formula comes in. For three-dimensional space, we calculate the distance between two points \[r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\]This equation helps us measure how far apart the charges are. It applies the Pythagorean theorem, extending it into three dimensions. In our example, we calculate the distances between charge A and the other charges B, C, and D, to find their pairwise separations.

Vector Addition

When multiple forces act on a point, we need to consider both magnitude and direction. This is where vector addition comes into play. Each force acting on charge A is represented as a vector. By calculating the vector sum, we determine the overall effect. In this exercise, since the forces due to C and D cancel each other out due to symmetry, the resultant force is only due to charge B. This highlights the importance of understanding vector directions and the net effect of balanced charges.

Force Calculation

Using Coulomb's Law, we can calculate the force between two charges:\[F = k \frac{q_1 q_2}{r^2}\]Here, \(k\) is the Coulomb's constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them. This equation tells us how strong the force is and in which direction it acts. In our example, calculating the forces between charge A and all other charges involves plugging into this formula. After vector addition, we find that only the force due to charge B impacts the total force on A. Understanding this calculation helps predict how charges interact in spaces like electric fields.