Two pith balls below each have a mass of 5.0 g and equal charge. One pith ball is suspended by an insulating thread. The other is brought to x = 4.0 cm from the suspended ball. The suspended ball is now hanging with the thread forming an angle of 30.0° with the vertical. The ball is in equilibrium with FE, Fg, and FT. Calculate each of the following.
Fg = 0.049
FE = 0.028
What is the charge on the pith balls?
Let me get it straight? You are given the electric force, and the distance, and you want the charge?
F=Kq^2/r^2 solve for q.
I got the answer as 1E-7, but it wouldn't accept the answer, any ideas?
To find the charge on the pith balls, we can first start by identifying the forces acting on the suspended ball.
1. Gravitational Force (Fg): The mass of the ball is given as 5.0 g. By multiplying it with the acceleration due to gravity (g = 9.8 m/s^2), we can find the gravitational force acting on the ball. The given value for Fg is 0.049 N, which represents the magnitude of the gravitational force.
2. Electrostatic Force (FE): The given value for FE is 0.028 N, which represents the magnitude of the electrostatic force.
3. Tension in the Thread (FT): The tension in the thread counteracts the gravitational force and creates an equilibrium condition for the ball. Since the ball is suspended and not moving, the tension in the thread must be equal to the gravitational force. Therefore, FT = Fg = 0.049 N.
Now, let's break down the gravitational force (Fg) into its vertical and horizontal components based on the angle formed by the thread with the vertical.
The vertical component of Fg can be determined using trigonometry. The vertical component is given by Fg_vertical = Fg * sin(30°).
Substituting the given value of Fg into the equation:
Fg_vertical = 0.049 N * sin(30°) = 0.025 N
Since the electrostatic force (FE) completely opposes the vertical component of the gravitational force (Fg_vertical), we can equate them: FE = Fg_vertical = 0.025 N.
Now, we can equate the electrostatic force (FE) to the Coulomb's law equation:
FE = k * (q1 * q2) / r^2
where k is the electrostatic constant (9.0 * 10^9 Nm^2/C^2), q1 is the charge on the suspended ball, q2 is the charge on the created ball, and r is the distance between them (4.0 cm = 0.04 m).
Rearranging the equation to solve for q1:
q1 = (FE * r^2) / (k * q2)
Substituting the given values:
q1 = (0.025 N * (0.04 m)^2) / (9.0 * 10^9 Nm^2/C^2 * q2)
To find q2, we need to know the charge on the created ball. Unfortunately, the charge on the created ball is not given in the question. Without that information, we cannot calculate the charge on the pith balls.