A spring with an unstrained length of 0.074 m and a spring constant of 2.4 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 5.1 X 10^�3 kg and a net charge of +�6.6 �C 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?
I believe that m= 5.1 X 10^- 3 kg
F(electr) + F(spr) = mg,
qE + k•ΔL=mg
E=( mg- k•ΔL)/q=
={5.1•10⁻³•9.8 – 2.4(0.074-0.059)}/6.6 =… V/m
To find the magnitude of the external electric field, we can use Hooke's Law and the equilibrium conditions.
Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. Mathematically, this can be written as:
F = -k * x
Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, since the spring hangs vertically downward, the weight of the sphere will create a force in the opposite direction of the electric field. At equilibrium, these two forces will be equal in magnitude.
The weight of the sphere can be calculated using the equation:
Weight = mass * gravity
Where mass is the mass of the sphere and gravity is the acceleration due to gravity.
Given:
Mass of the sphere (m) = 5.1 x 10^3 kg
Gravity (g) = 9.8 m/s^2
Weight = (5.1 x 10^3 kg) * (9.8 m/s^2)
Next, we can set up an equation for the forces at equilibrium:
-k * x = Weight
Rearranging this equation, we get:
-k * (equilibrium length - unstrained length) = Weight
Plugging in the given values:
-2.4 N/m * (0.059 m - 0.074 m) = Weight
Simplifying:
-0.036 N = Weight
Now, we have two forces that should cancel each other out:
Magnitude of the electric field (E) = Weight / charge
Given:
Charge (q) = 6.6 C
Now, let's calculate the magnitude of the electric field:
E = (-0.036 N) / (6.6 C)
The magnitude of the external electric field is approximately -0.0055 N/C (or 5.5 N/C) directed vertically upward.
To find the magnitude of the external electric field, we can use the concept of electrostatic equilibrium.
When an object is in electrostatic equilibrium, the electric force acting on it is balanced by another force, in this case, the force exerted by the spring. At equilibrium, these two forces are equal in magnitude and opposite in direction.
The force exerted by the spring is given by Hooke's law: F_spring = -k * Δx, where k is the spring constant and Δx is the change in length from the equilibrium position.
In this case, the change in length of the spring from the equilibrium position is given by Δx = x_eq - x_unstrained, where x_eq is the equilibrium length of the spring and x_unstrained is the unstrained length of the spring.
Using the given values, we have:
k = 2.4 N/m
x_eq = 0.059 m
x_unstrained = 0.074 m
Substituting these values into the equation, we get:
Δx = 0.059 m - 0.074 m = -0.015 m (negative because the spring is stretched)
Now, setting the force exerted by the spring equal to the electric force, we have:
-k * Δx = q * E
Where q is the charge on the sphere and E is the magnitude of the external electric field.
In this case, q = 6.6 C (positive because it is a net positive charge) and we need to solve for E.
Rearranging the equation, we have:
E = (-k * Δx) / q
Substituting the values we have:
E = (-2.4 N/m * -0.015 m) / 6.6 C
Calculating this, we get:
E = 0.00545 N/C
Therefore, the magnitude of the external electric field is approximately 0.00545 N/C.