A spring with an unstrained length of 0.075 m and a spring constant of 2.7 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 4.84 10-3 kg and a net charge of +7.0 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 0.061 m. What is the magnitude of the external electric field?
To find the magnitude of the external electric field, we can use the principle of equilibrium. At equilibrium, the gravitational force on the sphere due to its weight is balanced by the electrostatic force on the sphere due to the electric field.
Let's break down the forces acting on the sphere:
1. Weight of the sphere (gravitational force, Fg):
The weight of the sphere can be calculated using the formula:
Fg = m * g
where m is the mass of the sphere and g is the acceleration due to gravity. In this case, m = 4.84 x 10^(-3) kg and g = 9.8 m/s^2.
2. Electrostatic force on the sphere (Fe):
The electrostatic force on the sphere can be calculated using the formula:
Fe = q * E
where q is the net charge on the sphere and E is the magnitude of the external electric field. In this case, q = 7.0 x 10^(-6) C.
At equilibrium, the electrostatic force Fe must balance the gravitational force Fg. Therefore, we can set up the equation:
Fe = Fg
q * E = m * g
Solving this equation for E, we get:
E = (m * g) / q
Substituting the given values, we have:
E = ((4.84 x 10^(-3) kg) * (9.8 m/s^2)) / (7.0 x 10^(-6) C)
Evaluating this expression gives us the magnitude of the external electric field.