Final answer:
The magnitude of the electric force exerted by the tape on the paper at a distance of 8.0 mm is 3.72 x 10^-7 C. The direction of the force is repulsive as like charges repel each other.
Step-by-step explanation:
When the tape comes within 8.0 mm of the scrap of paper, the electric force magnitude is great enough to overcome the gravitational force exerted by Earth on the scrap and lift it. To determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance, we can use the equation for the electric force:
F = k * (Q1 * Q2) / r^2
where F is the magnitude of the force, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.
We know that the gravitational force is equal to the weight of the paper:
Fg = m * g
where Fg is the gravitational force, m is the mass of the paper (190 mg = 0.190 g), and g is the acceleration due to gravity (9.8 m/s^2).
At the point where the tape comes within 8.0 mm of the paper, the electric force is equal to the gravitational force:
F = Fg
k * (Q1 * Q2) / r^2 = m * g
We can solve this equation for the magnitude of the charge Q1 or Q2:
Q1 * Q2 = (m * g * r^2) / k
Substituting the known values, we get:
Q1 * Q2 = (0.190 g * 9.8 m/s^2 * (8.0 mm)^2) / (8.99 x 10^9 Nm^2/C^2)
Q1 * Q2 = 1.383 x 10^-12 C^2
To find the magnitude of the charge, we can assume that Q1 and Q2 are equal, so:
Q1 = Q2 = sqrt(1.383 x 10^-12 C^2) = 3.72 x 10^-7 C
The magnitude of the electric force exerted by the tape on the paper at this distance is 3.72 x 10^-7 C. The direction of the force is repulsive because like charges repel each other.