Plot a graph showing the variation of coulomb’s force (F) versus 1/r^2, where r is distance between the charges of each pair of charges: (1μC, 2 μC) and (2 μC, – 3 μC) . Interpret the graphs obtained

To plot a graph showing the variation of Coulomb's force (F) versus \( \frac{1}{r^2} \) for the given pairs of charges, we first need to understand how Coulomb's law works. Coulomb's law states that the force \( F \) between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance \( r \) between them. The formula can be expressed as:

F = k * (|q1 * q2|) / r^2

Here, \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \), and \( q1 \) and \( q2 \) are the magnitudes of the charges. Let's analyze the two pairs of charges you provided: (1 μC, 2 μC) and (2 μC, -3 μC).

Calculating the Forces

We will calculate the force for each pair of charges at different distances \( r \) and then plot \( F \) against \( \frac{1}{r^2} \).

Pair 1: (1 μC, 2 μC)

• Let \( q1 = 1 \times 10^{-6} \, \text{C} \)
• Let \( q2 = 2 \times 10^{-6} \, \text{C} \)

Using Coulomb's law:

F = k * (|1 μC * 2 μC|) / r^2 = (8.99 \times 10^9) * (2 \times 10^{-12}) / r^2

Now, let's calculate \( F \) for various values of \( r \):

• If \( r = 1 \, \text{m} \), then \( F = 17.98 \, \text{N} \)
• If \( r = 2 \, \text{m} \), then \( F = 4.495 \, \text{N} \)
• If \( r = 3 \, \text{m} \), then \( F = 2.0 \, \text{N} \)

Pair 2: (2 μC, -3 μC)

• Let \( q1 = 2 \times 10^{-6} \, \text{C} \)
• Let \( q2 = -3 \times 10^{-6} \, \text{C} \)

Again, applying Coulomb's law:

F = k * (|2 μC * -3 μC|) / r^2 = (8.99 \times 10^9) * (6 \times 10^{-12}) / r^2

Calculating \( F \) for various values of \( r \):

• If \( r = 1 \, \text{m} \), then \( F = 53.94 \, \text{N} \
• If \( r = 2 \, \text{m} \), then \( F = 13.485 \, \text{N} \
• If \( r = 3 \, \text{m} \), then \( F = 5.96 \, \text{N} \

Graphing the Results

Now that we have the values of \( F \) for different distances \( r \), we can calculate \( \frac{1}{r^2} \) for each distance:

• If \( r = 1 \, \text{m} \), \( \frac{1}{r^2} = 1 \)
• If \( r = 2 \, \text{m} \), \( \frac{1}{r^2} = 0.25 \)
• If \( r = 3 \, \text{m} \), \( \frac{1}{r^2} = 0.111 \)

Now, we can plot \( F \) against \( \frac{1}{r^2} \) for both pairs of charges. The x-axis will represent \( \frac{1}{r^2} \), and the y-axis will represent the force \( F \).

Interpreting the Graphs

When you plot the data, you will notice that both graphs will show a linear relationship between \( F \) and \( \frac{1}{r^2} \). This linearity indicates that as the distance between the charges decreases (which increases \( \frac{1}{r^2} \)), the force between the charges increases. The slope of each line will differ based on the product of the charges involved:

• The first pair (1 μC, 2 μC) will have a lower slope compared to the second pair (2 μC, -3 μC) because the product of the charges is smaller.
• The second pair will show a stronger force due to the larger product of the charges, indicating a more significant interaction.

This analysis not only reinforces the concept of Coulomb's law but also illustrates how the force between charged objects varies with distance, providing a clear visual representation of the relationship.