Electric Force of Attraction between Pennies

What is the relationship between the electrostatic force of attraction and the weight of a penny?

The electrostatic force of attraction between two pennies is equal to the weight of a penny. In what state are the pennies considered to be in equilibrium?

Answer:

The electrostatic force of attraction between two pennies is equal to the weight of a penny, indicating that they are in a state of equilibrium. The equilibrium position occurs when there is no net force acting on the separation between the two pennies.

When the electrostatic force of attraction between two pennies is equal to the weight of a penny, they are said to be in equilibrium. This equilibrium occurs when the forces due to the electrostatic attraction and the weight of the penny balance each other out, resulting in a stable separation between the pennies.

The relationship between the electrostatic force of attraction and the weight of a penny can be explained using Coulomb's law, which states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

In the case of two pennies, with equal and opposite charges, Coulomb's law can be written as F = kq^2/d^2, where F is the force, k is Coulomb's constant, q is the charge on each penny, and d is the distance between the charges.

By rearranging the equation and substituting the weight of the penny, we can calculate the separation between the two pennies using the formula d = sqrt(kq^2/mg). Plugging in the values for k, q, m (mass of the penny), and g (acceleration due to gravity), we find that the separation is 0.15 meters or 15 cm.

Understanding the equilibrium state between charges and gravitational forces can provide insights into the delicate balance of forces at play in seemingly simple phenomena.