When a charged balloon sticks to a wall, the downward gravitational force is balanced by an upward static friction force. The normal force is provided by the electrical attraction between the charged balloon and the equal but oppositely charged polarization induced in the wall's molecules.
If the mass of a balloon is 1.9 g, its coefficient of static friction with the wall is 0.70, and the average distance between the opposite charges is 0.80 mm, what minimum amount of charge must be placed on the balloon in order for it to stick to the wall?
Express your answer to two significant figures and include appropriate units.
To find the minimum amount of charge required for the balloon to stick to the wall, we need to consider the forces involved and apply the laws of static friction.
First, let's convert the mass of the balloon to kilograms:
1.9 g = 0.0019 kg
Next, we can calculate the gravitational force acting on the balloon:
F_gravity = mass * acceleration due to gravity
F_gravity = 0.0019 kg * 9.8 m/s^2
F_gravity ≈ 0.01862 N
According to the problem statement, the coefficient of static friction between the balloon and the wall is 0.70. The formula for static friction is given by:
F_friction = μ_s * F_normal
where F_friction is the frictional force, μ_s is the coefficient of static friction, and F_normal is the normal force. In this case, the normal force is provided by the electrical attraction between the balloon and the wall.
We can find the normal force by considering the electrical attraction:
F_electric = k * (q1 * q2) / r^2
where F_electric is the electrical force, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the opposite charges.
In this case, the electrical force is balanced by the gravitational force, so we have:
k * (q1 * q2) / r^2 = F_gravity
Rearranging the equation, we can solve for the magnitude of one of the charges:
q1 * q2 = (F_gravity * r^2) / k
Now consider that we only need the magnitude of the charge, so we can rewrite this equation as:
q^2 = (F_gravity * r^2) / k
Finally, we can solve for the magnitude of the charge q:
q = sqrt((F_gravity * r^2) / k)
Now, let's plug in the values and calculate the minimum amount of charge required:
q = sqrt((0.01862 N * (0.0008 m)^2) / (9 × 10^9 N m^2/C^2))
Calculating the value, we get approximately:
q ≈ 1.961 × 10^-6 C
Therefore, the minimum amount of charge required for the balloon to stick to the wall is approximately 1.961 × 10^-6 Coulombs.