An unstrained horizontal spring has a length of 0.33 m and a spring constant of 209 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.021 m relative to its unstrained length. Determine the possible algebraic signs and the magnitude of the charges.
1738
To determine the possible algebraic signs and the magnitude of the charges, we can use Hooke's Law and the electrostatic force equation.
First, let's find the force exerted by the spring using Hooke's Law:
F = k * x
F: force exerted by the spring
k: spring constant (209 N/m)
x: displacement from the unstrained position (0.021 m)
F = 209 N/m * 0.021 m
F = 4.389 N
Now, let's consider the electrostatic force between the two charges. The force exerted by each charge on the other is given by Coulomb's Law:
F = (k * |q1 * q2|) / r^2
F: electrostatic force
k: electrostatic constant (9 * 10^9 N*m^2/C^2)
q1, q2: charges on the objects
r: distance between the charges
Since the charges have equal magnitudes, let's represent the magnitude as q.
F = (k * |q * q|) / r^2
F = (k * q^2) / r^2
Comparing the equations for the force exerted by the spring and the electrostatic force, we can equate them:
4.389 N = (k * q^2) / r^2
Substituting the known values:
4.389 N = (209 N/m * q^2) / (0.33 m)^2
To find the possible algebraic signs of the charges, we consider the repulsive or attractive nature of the forces. Since the spring stretches, the charges must have the same sign, which means they are either both positive or both negative.
Now, let's solve for the magnitude of the charges:
4.389 N = (209 N/m * q^2) / (0.33 m)^2
Simplifying:
(209 N/m * q^2) = 4.389 N * (0.33 m)^2
209 N/m * q^2 = 4.389 N * 0.1089 m^2
q^2 = (4.389 N * 0.1089 m^2) / 209 N/m
q^2 = 0.00228 C^2
q = √(0.00228 C^2)
q ≈ ± 0.0477 C
Therefore, the possible algebraic signs of the charges are ± (positive or negative) and the magnitude of the charges is approximately 0.0477 C.