1.	Set q1 to 5C and fix the distance between the two objects at 6 cm, record them in table 2.

2.	Change the charge of object 2 as shown in table 2 and for each q2 record the electric force between the two objects.

3.	Plot a graph in Excel relating FE (y-axis) and q2 (x-axis). Sketch the graph or insert a screenshot.  Comment on the graph’s appearance.

4.	Find the slope of the graph and use it along with Coulomb’s Law to find the electric constant k.

5.	Calculate the percentage error for your measured value of k (known=8.99 × 109 Nm2/C2)
slope 12.483

To solve this problem, let's break down each part step by step.

### 1. Setting Up the Problem

- **Given:**
  - \( q_1 = 5 \, \mu C = 5 \times 10^{-6} \, C \)
  - Distance between the two objects \( r = 6 \, cm = 0.06 \, m \)

- **Table 2:** Let's assume it includes different values of \( q_2 \) and requires calculating the electric force \( F_E \) for each.

### 2. Calculating Electric Force \( F_E \)

According to Coulomb's Law, the electric force between two point charges is given by:
\[
F_E = \frac{k \cdot |q_1 \cdot q_2|}{r^2}
\]
where:
- \( k = 8.99 \times 10^9 \, N \cdot m^2/C^2 \) (Coulomb's constant)
- \( q_1 = 5 \times 10^{-6} \, C \)
- \( r = 0.06 \, m \)
- \( q_2 \) varies as per the table.

### 3. Plotting the Graph in Excel

To plot the graph:
1. Enter the values of \( q_2 \) in one column.
2. Calculate the electric force \( F_E \) for each \( q_2 \) and enter them in the adjacent column.
3. Create a scatter plot in Excel with \( q_2 \) on the x-axis and \( F_E \) on the y-axis.

The graph should be linear because \( F_E \) is directly proportional to \( q_2 \).

### 4. Finding the Slope and Calculating \( k \)

You’ve mentioned that the slope of the graph is \( 12.483 \).

From Coulomb's Law:
\[
\text{slope} = \frac{k \cdot q_1}{r^2}
\]
Rearranging for \( k \):
\[
k = \frac{\text{slope} \cdot r^2}{q_1}
\]
Substitute the given values:
\[
k = \frac{12.483 \cdot (0.06)^2}{5 \times 10^{-6}}
\]
Let's calculate this:

\[
k = \frac{12.483 \cdot 0.0036}{5 \times 10^{-6}} = \frac{0.0449388}{5 \times 10^{-6}} = 8.98776 \times 10^9 \, N \cdot m^2/C^2
\]

### 5. Calculating the Percentage Error

Percentage error is calculated using:
\[
\text{Percentage Error} = \left|\frac{\text{Experimental } k - \text{Theoretical } k}{\text{Theoretical } k}\right| \times 100\%
\]
Substitute the known values:
\[
\text{Percentage Error} = \left|\frac{8.98776 \times 10^9 - 8.99 \times 10^9}{8.99 \times 10^9}\right| \times 100\% \approx 0.027\%
\]

### Comment on the Graph
The graph of \( F_E \) versus \( q_2 \) should be a straight line passing through the origin, indicating a direct proportionality between the electric force and the magnitude of charge \( q_2 \). The linearity confirms Coulomb's law.

Would you like the details of the graph, or do you have any questions?

### Additional Questions:
1. How would the graph change if the distance between the charges was doubled?
2. What would happen to the slope if \( q_1 \) were increased?
3. How would the graph look if \( q_1 \) and \( q_2 \) had opposite signs?
4. How does the electric force between the charges change if both charges are doubled?
5. What assumptions are made in Coulomb's Law that might affect the accuracy of real-world measurements?

**Tip:** Always double-check unit conversions, especially when dealing with micro or nano quantities in physics.

Explore the concept of electric force using Coulomb's Law in physics. Solve problems involving electric charges, distances, and forces, and analyze relationships through graphical representation. Suitable for advanced high school students.

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