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 need to equate the forces acting on the charged sphere attached to the spring.
The force due to gravity acting on the sphere is given by the formula:
F_gravity = m * g
where m is the mass of the sphere and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = (5.09 * 10^-3 kg) * (9.8 m/s^2)
F_gravity = 4.98 * 10^-2 N
The force exerted by the spring is given by Hooke's Law:
F_spring = k * (change in length)
where k is the spring constant and (change in length) is the difference between the equilibrium length and the unstrained length of the spring.
F_spring = 2.6 N/m * (0.059 m - 0.076 m)
F_spring = -0.0444 N
Since the spring is hanging vertically downward and the electric field is directed vertically upward, the net vertical force should be zero at equilibrium.
Net force = F_gravity + F_spring = 0
Therefore:
4.98 * 10^-2 N + (-0.0444 N) = 0
To balance this net force, the electric field must create an electric force on the charged sphere equivalent to the gravitational and spring forces.
Let's calculate the electric force with the formula:
F_electric = q * E
where q is the charge of the sphere and E is the electric field strength.
F_electric = (6.5 * 10^-6 C) * E
Since the net force must be zero, we can set the electric force equal to the sum of the gravitational and spring forces:
(6.5 * 10^-6 C) * E = 4.98 * 10^-2 N + (-0.0444 N)
Now we can solve for the electric field strength, E:
E = (4.98 * 10^-2 N + (-0.0444 N)) / (6.5 * 10^-6 C)
E = 7.2 * 10^3 N/C
Therefore, the magnitude of the external electric field is 7,200 N/C.
To find the magnitude of the external electric field, we can use Hooke's law for springs in equilibrium and the force exerted by the electric field.
1. Let's first calculate the force exerted by the spring using Hooke's law:
The spring constant, k = 2.6 N/m
Equilibrium length of the spring, L_eq = 0.059 m
Unstrained length of the spring, L_unstrained = 0.076 m
The displacement of the spring, ΔL = L_eq - L_unstrained
= 0.059 m - 0.076 m
= -0.017 m (Note: Negative sign indicates elongation)
The force exerted by the spring, F_spring = -k * ΔL
2. Next, we need to consider the force exerted by the electric field, which balances out the force by the spring for equilibrium.
The electric field force, F_electric = q * E
Mass of the sphere, m = 5.09 * 10^-3 kg
Net charge of the sphere, q = +6.5 * 10^-6 C
Gravitational force acting on the sphere, F_gravity = m * g (where g is the acceleration due to gravity)
At equilibrium, the spring force (F_spring) and the electric force (F_electric) balance each other, and thus, the net force on the sphere is zero.
Therefore, F_spring + F_gravity = F_electric
3. We can now substitute the values into the equation to solve for the external electric field:
-k * ΔL + m * g = q * E
Rearranging the equation, we can solve for the electric field (E):
E = (-k * ΔL + m * g) / q
Plugging in the known values:
E = (-2.6 N/m * -0.017 m + 5.09 * 10^-3 kg * 9.8 m/s^2) / 6.5 * 10^-6 C
Calculating the magnitude of the external electric field will yield the answer in N/C.