Calculating the Force of Attraction Between Two Point Charges Using Coulomb’s Law

Understanding the behavior of charges is fundamental in physics. One of the most common applications involves calculating the force of attraction or repulsion between two point charges. This guide will walk you through the process using Coulomb’s Law. Specifically, we'll delve into a real-world example to apply the formula accurately.

Understanding Coulomb’s Law

Coulomb’s Law, formulated by French physicist Charles Augustin de Coulomb, is a key principle in electromagnetism, describing the force between two point charges. The law states that the force of attraction or repulsion between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, this is represented as:

F k frac{q_1 q_2}{r^2}

Where:

F is the magnitude of the force between the charges. k is Coulomb’s constant, approximately 8.99 times 10^9 , text{Nm}^2/text{C}^2. q_1 and q_2 are the magnitudes of the charges in coulombs (C). r is the distance between the charges in meters (m).

Problem Context

Consider two point charges: a charge of 3 , mutext{C} located at (-2, 3) , text{m} and a charge of -1.2 , mutext{C} located at (0, 5) , text{m} on the x-y plane. We aim to find the magnitude and direction of the force of attraction between these charges.

Step-by-Step Solution

Step 1: Identifying the Charges and Their Positions

The two charges are:

q_1 3 , mutext{C} 3 times 10^{-6} , text{C} at location (-2, 3) , text{m}. q_2 -1.2 , mutext{C} -1.2 times 10^{-6} , text{C} at location (0, 5) , text{m}.

Step 2: Calculating the Distance r Between the Charges

To find the distance r, we use the distance formula:

(r sqrt{(y_2 - y_1)^2 (x_2 - x_1)^2})

Plugging in the coordinates:

(r sqrt{(5 - 3)^2 (0 - (-2))^2} sqrt{2^2 2^2} sqrt{8} 2sqrt{2} , text{m})

Step 3: Calculating the Magnitude of the Force F

Substituting the values into Coulomb’s Law:

(F 8.99 times 10^9 frac{3 times 10^{-6} times -1.2 times 10^{-6}}{(2sqrt{2})^2})

First, calculate (2sqrt{2})^2:

((2sqrt{2})^2 8)

Substitute this value into the equation:

(F 8.99 times 10^9 frac{3 times 10^{-6} times -1.2 times 10^{-6}}{8} 8.99 times 10^9 frac{-3.6 times 10^{-12}}{8})

Calculate the force:

(F 8.99 times 10^9 times -4.5 times 10^{-13} -4.0455 times 10^{-3} , text{N} approx -4.05 , text{mN})

The magnitude of the force of attraction is approximately 4.05 , text{mN}. The negative sign indicates that the force is attractive.

Step 4: Determining the Direction of the Force

The force between the two charges acts along the line connecting the charges. Since one charge is positive and the other is negative, the force is attractive.

Step 5: Calculating the Unit Vector from q_1 to q_2

The vector pointing from q_1 to q_2 is:

(vec{r} 2, 2)

The magnitude of this vector is:

(|vec{r}| sqrt{2^2 2^2} 2sqrt{2})

The unit vector (hat{r}) in the direction from q_1 to q_2 is:

(hat{r} left( frac{2}{2sqrt{2}}, frac{2}{2sqrt{2}} right) left( frac{1}{sqrt{2}}, frac{1}{sqrt{2}} right))

Final Result

The magnitude of the force of attraction is approximately 4.05 , text{mN}, and the direction is along the vector (left( frac{1}{sqrt{2}}, frac{1}{sqrt{2}} right)) or approximately 45^circ above the positive x-axis.

In conclusion, the force of attraction between the two charges is approximately 4.05 , text{mN} directed at an angle of 45^circ from the positive x-axis towards the second charge.

Additional Resources

For further study, you can explore additional resources on Coulomb’s Law and other fundamental principles of electromagnetism. Understanding these concepts will help you solve a wide range of physics problems.