Question

An unstrained horizontal spring has a length of 0.26 m and a spring constant of 180 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 0.039 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.

Answer 1

The charges on the objects have equal magnitudes and must be of the same sign (either both positive or both negative) because the spring is stretched due to repulsion. The force exerted by the spring can be calculated using Hooke's law and set equal to the electric force between the charges to solve for their magnitude.

Step-by-step explanation:

To determine the magnitude of the charges on two small charged objects attached to a spring, we must use Hooke's law and Coulomb's law. Due to the charges, the spring stretches by 0.039 m, which results in an additional force that can be calculated with Hooke's law (F = kx), where k is the spring constant, and x is the extension.

Assuming the charges are the same, and that the objects repel each other (since the spring lengthens), they must have like charges, either both positive or both negative. This is the answer to part (a).

For part (b), using Hooke's law, the force exerted by the spring at this stretch is F = kx = 180 N/m * 0.039 m = 7.02 N. Since this force is due to the electric repulsion between the charges, we can set it equal to the Coulomb force equation, F = k_e * q^2 / r^2, and solve for q, the charge's magnitude. Here, k_e is Coulomb's constant (approximately 8.99 × 10^9 N m^2/C^2), and r is the separation between the charges, which is the length of the spring when stretched: 0.26 m + 0.039 m = 0.299 m.

By equating Hooke's law to Coulomb's law, we can find the charges have a magnitude of approximately 7.02 N = (8.99 × 10^9 N m^2/C^2) * q^2 / (0.299 m)^2. Solving for q gives us the magnitude of the charges.

Answer 2

a)

two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)

b)

both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

Step-by-step explanation:

Given that;

L = 0.26 m

k = 180 N/m

x = 0.039 m

a)

we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.

b)

Spring force F = kx

F = 180 × 0.039

F = 7.02 N

Now, Electrostatic force F = Keq²/r²

where r = L + x = ( 0.26 + 0.039 )

we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2

so from the equation; F = Keq²/r²

Fr² = Keq²

q = √ ( Fr² / Ke )

we substitute

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ (0.627595 / 9×10⁹)

q = √(6.97 × 10⁻¹¹)

q = 8.35 × 10⁻⁶ C

Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C