Three identical small Styrofoam balls of mass
1.53 g are suspended from a fixed point by
three nonconducting threads, each with a
length of 56.4 cm and negligible mass. At
equilibrium the three balls form an equilateral triangle with sides of 31 cm.
What is the common charge carried by
each ball? The acceleration of gravity is
9.8 m/s
2
and the value of Coulomb’s constant
is 8.98755 × 10
9
N · m2
/C
2
.
Answer in units of µC
To determine the common charge carried by each ball, we can consider the forces acting on the balls in equilibrium.
First, let's calculate the mass of each ball in kg:
mass = 1.53 g = 0.00153 kg
The force of gravity acting on each ball is given by:
F_gravity = mass * acceleration due to gravity
F_gravity = 0.00153 kg * 9.8 m/s^2
F_gravity = 0.014994 N
Since the balls are in equilibrium, the tension in each thread must balance out the gravitational force acting on each ball.
Let's assume the common charge on each ball is Q (in Coulombs). The electrostatic force between two charged objects can be calculated using Coulomb's law:
F_electrostatic = (k * Q^2) / r^2
Where:
- k is Coulomb's constant: k = 8.98755 × 10^9 N * m^2 / C^2
- Q is the common charge on each ball in Coulombs
- r is the distance between the centers of the balls
In an equilateral triangle, the distance between the centers of the balls is equal to the side length divided by the square root of 3:
r = side length / √3
r = 31 cm / √3 = 17.88 cm = 0.1788 m
The electrostatic force acting on each ball can be split into horizontal and vertical components. Since the balls are in equilibrium, the vertical components of the electrostatic forces must balance out the vertical gravitational force:
F_electrostatic_vertical = 2 * (F_electrostatic * sin(60 degrees))
Setting this equal to the gravitational force:
2 * (F_electrostatic * sin(60 degrees)) = F_gravity
Now, let's solve for the common charge carried by each ball:
F_electrostatic = F_gravity / (2 * sin(60 degrees))
(Q^2 * k) / r^2 = F_gravity / (2 * sin(60 degrees))
Q^2 * k = (F_gravity * r^2) / (2 * sin(60 degrees))
Q^2 = ((F_gravity * r^2) / (2 * sin(60 degrees))) / k
Q^2 = (0.014994 N * (0.1788 m)^2) / (2 * sin(60 degrees) * 8.98755 × 10^9 N * m^2 / C^2)
Q^2 ≈ 2.563 × 10^-11 C^2
Taking the square root to find the value of Q:
Q ≈ √(2.563 × 10^-11 C^2)
Q ≈ 1.601 × 10^-6 C
Finally, we need to express the answer in units of µC, which stands for microCoulombs.
1 µC = 10^-6 C
Converting the value of Q to µC:
Q ≈ 1.601 × 10^-6 C * (1 µC / 10^-6 C)
Q ≈ 1.601 µC
Therefore, the common charge carried by each ball is approximately 1.601 µC.