91Ó°ÊÓ

Electrostatic force is the force of attraction or repulsion between two charged objects. It follows an inverse square law, meaning the force decreases as the square of the distance between the objects increases. In our exercise, we use Coulomb's Law to calculate this force. Coulomb's Law is given by the formula: \[ F = k_e \frac{|q_1 q_2|}{r^2} \] Here,

• F is the electrostatic force.
• \(k_e\) (Coulomb's constant) is approximately \(8.99 \times 10^9 \text{N} \text{m}^2/\text{C}^2\).
• |\(q_1 q_2\)| are the magnitudes of the charges.
• r is the distance between the charges.

To find the force between our water drops, we substitute the values given: \[ F = 8.99 \times 10^9 \frac{(1.00 \times 10^{-16} \text{C})^2}{(1.00 \times 10^{-2} \text{m})^2} \] After calculation, the result is: \[ F = 8.99 \times 10^{-21} \text{N} \].This is a very small force, typical for interactions at the microscopic level.

Charge imbalance occurs when an object has more or fewer electrons than protons. This imbalance results in a net electric charge. In the context of our exercise, each water drop has a charge of \(-1.00 \times 10^{-16} \text{C}\). Since electrons carry a negative charge, having an excess of electrons means the water drops are negatively charged. We can determine the total number of these excess electrons by dividing the total charge of the water drop by the charge of a single electron. The charge of a single electron is \(-1.60 \times 10^{-19} \text{C}\). To find the number of excess electrons on a drop, we use the following calculation:\[ \frac{-1.00 \times 10^{-16} \text{C}}{-1.60 \times 10^{-19} \text{C/electron}} = 625 \].There are 625 excess electrons on each water drop.

Excess electrons refer to the additional electrons present on an object, giving it a negative charge. Each excess electron contributes \(-1.60 \times 10^{-19} \text{C}\) of charge. When substances, like our tiny water drops, acquire extra electrons, they become negatively charged. To determine how many electrons make up the net charge, we divide the total measured charge by the charge of one electron. From the exercise, each water drop carries a total charge of \(-1.00 \times 10^{-16} \text{C}\). Using the charge of a single electron, we calculate the number of excess electrons as follows:\[ \frac{-1.00 \times 10^{-16} \text{C}}{-1.60 \times 10^{-19} \text{C/electron}} = 625 \].Therefore, each water drop possesses 625 excess electrons, making up its negative charge imbalance.