A 3.0 g ping pong ball is suspended from a thread 35 cm long. When a comb is brought to the same height, the ping pong ball is repelled and the thread makes an angle of 10.0o with the vertical. What is the electric force exerted on the ping pong ball?
m•g = T•sinα
F(el) =T•cosα
m•g/F(el) = sinα/cosα = tanα,
F(el) = m•g/tanα .
To determine the electric force exerted on the ping pong ball, we need to consider the forces acting on it.
Given:
Mass of the ping pong ball, m = 3.0 g = 0.003 kg
Length of the thread, L = 35 cm = 0.35 m
Angle made by the thread with the vertical, θ = 10.0°
First, let's determine the tension in the thread using the following steps:
1. Calculate the weight of the ping pong ball, which is given by the formula: weight (w) = mass (m) × acceleration due to gravity (g).
Here, g is approximately 9.8 m/s^2.
So, w = 0.003 kg × 9.8 m/s^2 = 0.0294 N.
2. The force causing the tension in the thread is the vertical component of the tension. This vertical component can be calculated using the formula: tension (T) = weight (w) × cos(θ).
Here, θ is the angle made by the thread with the vertical.
So, T = 0.0294 N × cos(10.0°) = 0.0294 N × 0.9848 ≈ 0.0289 N.
Next, let's determine the electric force (F) exerted on the ping pong ball using the following steps:
1. The electric force (F) can be found using Coulomb's Law:
F = k × (|q1| × |q2|) / r^2,
where k is the electrostatic constant (8.99 × 10^9 Nm^2/C^2),
|q1| and |q2| are the magnitudes of the charges on the objects, and
r is the distance between the charges.
2. In this case, the electric force (F) is equal in magnitude and opposite in direction to the tension in the thread. So, F = T.
Therefore, the electric force exerted on the ping pong ball is approximately 0.0289 N.