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
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
There are three forces, kx the spring
mg gravity
and qE the field force
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 ร 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 ร 10^(-6) C (given in the question)
2. Plug in these values into the equation:
(5.09 ร 10^(-3) kg) * (9.8 m/s^2) + (2.6 N/m) * (0.059 m) = (6.5 ร 10^(-6) C) * E
3. Simplify the equation:
0.049882 N + 0.1534 N = 6.5 ร 10^(-6) C * E
0.203282 N = 6.5 ร 10^(-6) C * E
4. Solve for E:
E = 0.203282 N / (6.5 ร 10^(-6) C)
5. Calculate the magnitude of the external electric field:
E โ 3.126 ร 10^4 N/C
Therefore, the magnitude of the external electric field is approximately 3.126 ร 10^4 N/C.