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.
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%= ...%

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