Use this Example calculation above, and the trendline equation (3) coefficients that you have obtained on your graph, to find k.  Show your work below. Be careful with powers of ten.

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Chapter1: Units, Trigonometry. And Vectors
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       F=0.8975r^-2.00

And re-writing it as:

        (4)F=0.8975/r^-2.00

 

 

We can now compare it with Coulomb’s equation: 

 

     (1)F=K|q1q2|/r^2

 

 

So, comparing these equations (4) and (1) we realize that:

k=|q1q2|=0.8975

 

Plugging in values for charges: q1 = -10µC, and q2 = 10µC, isolating  k (Coulomb’s constant), and recalling that 1µC = 10-6 C, we get:

k|-10µC*10µC|=0.8975

k=0.8934/|-10µC*10µC|= 0.8934/10*10^-6C*10*10^-6C (3)

 

Use this Example calculation above, and the trendline equation (3) coefficients that you have obtained on your graph, to find k.  Show your work below. Be careful with powers of ten. 

 

 

k=  0.8975/|-10µC*10µC|= 0.8975/10*10^-6C*10*10^-6C =     (5)

 

 

 

Now, important part in reporting the answer is properly showing its units. To derive the SI units for Coulomb’s constant k, let’s look again at the equation (1), and re-write it as:

 

    k=F*r^2/|q1q2| (6)

 

 

Recalling the units for all parameters in this equation:

Force, F – Newtons, [N]

Distance, r – meters, [m]

Electric charge, q – Coulomb, [C]

 

So, we get the following units for Coulomb’s constant k:

   k:[N*m^2/C^2]  (7)

 

Now, recalling Newton’s 2nd law from Physics 1:

F= m*a

    

 

Where a is acceleration, with units [m/s^2] , and m is mass with units [kg].

 

 

We can express the force units Newtons as:

    Force, F: [N]= [kg*m\s^2] (8)

 

Now, plugging equation (8) into (7), we obtain the final expression of units for Coulomb’s constant in SI units:

 

 

    k=kg*m*m^2/s^2*C^2= kg*m^3//s^2*C^2 (9)

 

 

So, you may now report your experimental value for Coulomb’s constant from (5):

 

 

Coulomb’s constant, your value:   kexperimental= ... kg*m^3/s^2*C^2

 

 

 

Step 3: Verifying your value and calculating percent error

 

 

Write down Coulomb’s constant accepted value (look for it in the corresponding textbook chapter) below:

kaccepted= ... kg*m^3/s^2*c^2

Use this accepted value and kexperimental you have obtained (5) to calculate the percentage error using the formula below. Show your work:

 

%error= |experimental value- accepted value|/accepted value *100%= ...%

       

 

## Coulomb's Law: Two Point Charges

### Description:
The graph above demonstrates Coulomb's Law for two point charges. The graph plots the relationship between the distance (in meters) and the force between the charges. The data points appear to follow an inverse square law, consistent with Coulomb's Law, which states that the magnitude of the electrostatic force 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 between them.

### Graph Explanation:
- **Title**: "Coulomb's Law, Two Point Charges"
- **X-Axis**: Distance, labeled as "Distance, m" ranging from 0 to 0.12 meters.
- **Y-Axis**: Axis Title (should typically represent force or electrostatic force) ranging from 0 to 2500 units.
 
### Observations:
- The plotted data points (shown in blue) are fit to a curve following the equation \( y = 0.8975x^{-2} \) with an \( R^2 \) value of 1, showing a perfect fit to the inverse square law.
- The force decreases rapidly as the distance increases, illustrating the inverse square relationship (force is inversely proportional to the square of the distance).

### Key Points:
- **Mathematical Representation**: The equation of the trendline is \( y = 0.8975x^{-2} \), where \( y \) represents the force, and \( x \) represents the distance.
- ** \( R^2 \) Value**: The coefficient of determination \( R^2 = 1 \) indicates a perfect correlation between the distance and the force in this context, adhering precisely to Coulomb’s Law.

This graph provides a visual representation of how the electrostatic force between two point charges diminishes as the distance between them increases, showcasing the fundamental principle of Coulomb's Law in physics.
Transcribed Image Text:## Coulomb's Law: Two Point Charges ### Description: The graph above demonstrates Coulomb's Law for two point charges. The graph plots the relationship between the distance (in meters) and the force between the charges. The data points appear to follow an inverse square law, consistent with Coulomb's Law, which states that the magnitude of the electrostatic force 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 between them. ### Graph Explanation: - **Title**: "Coulomb's Law, Two Point Charges" - **X-Axis**: Distance, labeled as "Distance, m" ranging from 0 to 0.12 meters. - **Y-Axis**: Axis Title (should typically represent force or electrostatic force) ranging from 0 to 2500 units. ### Observations: - The plotted data points (shown in blue) are fit to a curve following the equation \( y = 0.8975x^{-2} \) with an \( R^2 \) value of 1, showing a perfect fit to the inverse square law. - The force decreases rapidly as the distance increases, illustrating the inverse square relationship (force is inversely proportional to the square of the distance). ### Key Points: - **Mathematical Representation**: The equation of the trendline is \( y = 0.8975x^{-2} \), where \( y \) represents the force, and \( x \) represents the distance. - ** \( R^2 \) Value**: The coefficient of determination \( R^2 = 1 \) indicates a perfect correlation between the distance and the force in this context, adhering precisely to Coulomb’s Law. This graph provides a visual representation of how the electrostatic force between two point charges diminishes as the distance between them increases, showcasing the fundamental principle of Coulomb's Law in physics.
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