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
Note: the way my teacher told me to solve it is to use the equation: mg+ kx=qE. I know i have to solve for E but I don't know what to plug in for (g). Any help would be great! Thanks
Youre summing forces so mg is force due to gravity g=9.8
There are three forces, kx the spring
mg gravity
and qE the field force
To solve for the magnitude of the external electric field (E), we can use the equation provided by your teacher:
mg + kx = qE
Here's how you can solve for E:
1. Determine the values for the variables in the equation:
- m = mass of the sphere = 5.09 × 10^(-3) kg
- g = acceleration due to gravity = 9.8 m/s^2 (assuming we are on Earth)
- k = spring constant = 2.6 N/m
- x = displacement from the unstrained length to the equilibrium position = 0.059 m (given in the question)
- q = net charge on the sphere = +6.5 × 10^(-6) C (given in the question)
2. Plug in these values into the equation:
(5.09 × 10^(-3) kg) * (9.8 m/s^2) + (2.6 N/m) * (0.059 m) = (6.5 × 10^(-6) C) * E
3. Simplify the equation:
0.049882 N + 0.1534 N = 6.5 × 10^(-6) C * E
0.203282 N = 6.5 × 10^(-6) C * E
4. Solve for E:
E = 0.203282 N / (6.5 × 10^(-6) C)
5. Calculate the magnitude of the external electric field:
E ≈ 3.126 × 10^4 N/C
Therefore, the magnitude of the external electric field is approximately 3.126 × 10^4 N/C.