Calculation of Charges based on Repulsion Force

Given:
- Force of repulsion between two similar charges, F = 44.1N
- Distance between the charges, r = 2 cm

Step 1:
Determine the Coulomb's Law constant, k
- Coulomb's Law states that F = k(q1q2/r^2), where k = 9 x 10^9 Nm^2/C^2
- Substitute the given values:
- F = 44.1N
- r = 2 cm = 0.02 m
- Solve for k:
- k = F(r^2)/(q1q2) = (44.1N)(0.02m)^2/(q1q2) = 9 x 10^9 Nm^2/C^2
- k = 1.764 x 10^-11 C^2/Nm^2

Step 2:
Solve for the charges, q1 and q2
- Since the charges are similar, q1 = q2 = q
- Substitute the given values and k:
- F = k(q^2/r^2) = (1.764 x 10^-11 C^2/Nm^2)(q^2)/(0.02m)^2 = 44.1N
- Solve for q:
- q^2 = (44.1N)(0.02m)^2/(1.764 x 10^-11 C^2/Nm^2)
- q^2 = 2 x 10^-7 C^2
- q = ± 1.414 x 10^-4 C

Step 3:
Interpretation of the result
- Since the charges are similar, they must have the same magnitude but opposite signs to cause repulsion.
- Therefore, the charges are q1 = q2 = ± 1.414 x 10^-4 C.
- The negative sign indicates that the charges are electrons (negative charges).
- The charges are relatively small compared to typical charges in everyday objects.

Conclusion:
The magnitude of each charge that causes a repulsion force of 44.1N at a distance of 2 cm in air is ± 1.414 x 10^-4 C, which are electrons with opposite signs. Coulomb's Law constant, k, is 9 x 10^9 Nm^2/C^2.

F=q^2/r^2 (since the charges are similar )
44.1=q^2/(0.02)^2 (where q is charge)
44.1= q^2/(4×10^-4)
q^2=44.1×4×10^-4
q^2=1.764
q=1.328
the value of similar charges are 1.328 coulomb