Electric field/charge woed problem

[Dx]

A copper peny has a mass 3g. A total of 4 x 10^12 electrons are transferred from one neutral penny to another. If the lectrostatic force of attraction between the pennies is equal to the weight of a penny, what is the separatio between them?

what do i do here, i donno where to begin?

Dx

 

[quantumdude]

The problem pretty much tells you what to do. You are told the charge on each penny, so you should be able to set up the expression for the electric force between them (the separation r is unknown). You are also told the mass of a penny, so you should be able to calculate the weight.

Finally, you are told that the electric force is equal to the weight. Set them equal and solve for r.

 

[M98Ranger]

Don't worry, I can help you with this problem! First, we need to understand the equation for electrostatic force, which is F = k * (q1 * q2) / d^2, where k is the Coulomb's constant (9 * 10^9), q1 and q2 are the charges of the objects, and d is the distance between them. We also know that the weight of a penny is equal to its mass (3g) multiplied by the gravitational acceleration (9.8 m/s^2).

Since the pennies are neutral before the transfer of electrons, we can assume that the charge of each penny is 0. After the transfer, one penny will have a positive charge and the other will have a negative charge. Since we know that 4 x 10^12 electrons were transferred, we can calculate the charge of each penny using the formula q = ne, where n is the number of electrons and e is the elementary charge (1.6 * 10^-19 C).

So, the charge of each penny will be 6.4 * 10^-7 C. Now, we can equate the electrostatic force of attraction to the weight of a penny and solve for the distance between the pennies (d).

F = k * (q1 * q2) / d^2

F = m * g (weight of a penny)

k * (q1 * q2) / d^2 = m * g

Substituting the values,

9 * 10^9 * (6.4 * 10^-7)^2 / d^2 = 3 * 10^-3 * 9.8

Solving for d, we get d = 0.0013 m or 1.3 mm. This is the separation between the two pennies after the transfer of electrons.

I hope this helps you understand the problem and how to approach it. Let me know if you have any further questions. Good luck!