A small, m = 2.00 g plastic ball is suspended by a L = 25.0 cm long string in a uniform electric field, as shown in the figure below. If the ball is in equilibrium when the string makes a 14.0° angle with the vertical as indicated, what is the net charge on the ball?
I already answered the question. Thanks anyway
To determine the net charge on the plastic ball, we can use the concept of electrostatic equilibrium. In this situation, the electric force acting on the ball due to the electric field is balanced by the force of gravity pulling the ball downward.
To find the net charge, we need to note the following:
1. The weight of the ball is given by the equation:
w = mg
Where, m = mass of the ball = 2.00 g (given)
g = acceleration due to gravity = 9.8 m/s² (approximate value)
2. The force on the ball due to the electric field is given by:
Fe = qE
Where, q = charge on the ball (what we need to find)
E = magnitude of the electric field
3. The tension in the string can be resolved into two components: one acting horizontally and another vertically. The vertical component is equal to the weight of the ball, and the horizontal component is equal to the electrostatic force.
Now, let's analyze the forces acting on the ball:
1. The vertical component of the tension in the string is responsible for balancing the weight of the ball. It can be calculated using trigonometry:
Tsinθ = mg
Where, T = tension in the string
θ = angle made by the string with the vertical (given as 14.0°)
2. The horizontal component of the tension in the string is equal to the electrostatic force:
Tcosθ = qE
Since the string is in equilibrium, the tension remains constant. Therefore, we can equate the equations for the vertical and horizontal components of tension:
Tsinθ = Tcosθ
Now, we can solve for q using these equations:
Substituting T = mg/sinθ in Tcosθ = qE:
(mg/sinθ) cosθ = qE
Now, rearrange the equation to isolate q:
q = (mg cosθ) / (E sinθ)
Plug in the known values to find the net charge on the ball.