A spring with an unstrained length of 0.076 m and a spring constant of 2.6 N/m hangs vertically downward from the ceiling. A uniform electric field directed vertically upward fills the region containing the spring. A sphere with a mass of 5.09 10-3 kg and a net charge of +6.5 µC is attached to the lower end of the spring. The spring is released slowly, until it reaches equilibrium. The equilibrium length of the spring is 0.059 m. What is the magnitude of the external electric field?

Answer in N/C

To find the magnitude of the external electric field, we can first calculate the gravitational force acting on the sphere and then equate it to the electrical force on the charged sphere.

1. Begin by finding the gravitational force acting on the sphere.
The gravitational force is given by the equation F_g = m * g, where
- F_g is the gravitational force,
- m is the mass of the sphere,
- g is the acceleration due to gravity (approximately 9.8 m/s^2).

Plugging in the given values:
F_g = (5.09 * 10^-3 kg) * (9.8 m/s^2)
F_g ≈ 0.0499 N

2. Next, calculate the electrical force acting on the charged sphere.
The electrical force is given by the equation F_e = q * E, where
- F_e is the electrical force,
- q is the charge on the sphere,
- E is the magnitude of the electric field.

Plugging in the given values:
F_e = (6.5 * 10^-6 C) * E

3. Since the spring is at equilibrium, the gravitational force and electrical force are equal. Therefore, we can set up the equation F_g = F_e:
0.0499 N = (6.5 * 10^-6 C) * E

4. Solve the equation for E to find the magnitude of the external electric field:
E = (0.0499 N) / (6.5 * 10^-6 C)
E ≈ 7.68 * 10^3 N/C

Therefore, the magnitude of the external electric field is approximately 7.68 * 10^3 N/C.