91Ó°ÊÓ

Coulomb's Law is a fundamental principle in electrostatics, which describes how electric charges interact. It posits that the electric force between two point charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them. This is mathematically expressed as:\[F = k \frac{|Q_1 Q_2|}{r^2}\]where:

• \(F\) is the magnitude of the force between the charges,
• \(k\) is Coulomb's constant \, (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2),
• \(Q_1\) and \(Q_2\) are the amounts of the charges,
• \(r\) is the distance between the centers of the two charges.

In the context of the raindrop example, although we used the formula for electric potential, which also relies on Coulomb's constant, it's helpful to understand that the electric forces resulting from charge distributions are based on this law. The potential at the surface of the raindrop was calculated considering the cumulative effect of tiny imaginary charges through the volume of the drop, interacting with any charge placed at the surface.

Charge distribution refers to how electric charge is spread out in a given space or object. In our case of the raindrop, the charge is distributed uniformly, meaning the charge per unit volume is the same throughout the raindrop's volume.
When charges are distributed, both the shape and size of the object influence the electric potential and field they generate around them. The electric potential due to a uniformly charged sphere, like our raindrop, can conveniently be treated as if all its charge were concentrated at its center.
To calculate the electric potential on the surface of the raindrop, we used:\[V = \frac{kQ}{r}\]where:

• \(V\) is the electric potential,
• \(Q\) is the total charge of the sphere,
• \(r\) is the radius of the sphere.

For two raindrops merging, the new charge is the sum of their charges. Even though they merge into a larger drop, the charges still distribute uniformly. Thus, the larger raindrop's surface potential is influenced by its increased total charge and altered radius in the same way as a single sphere, as shown by the adjusted formula and calculations.

When two raindrops collide, their volumes and charges combine. The collision and merging are examples of a physical concept called 'conservation of mass and charge.'
From geometry, the volume \(V\) of a sphere is calculated using:\[V = \frac{4}{3}\pi r^3\]If you double the volume because two identical spheres merge, you need to recalculate the sphere's radius after invoking the formula.
Thus, if two raindrops with identical volume and charge specifications merge:

• Their new radius \(R\) follows from: \(R^3 = 2 \cdot (0.650\times 10^{-3})^3\).
• The charge combines to \(-2.40\) pC.

Calculating the larger raindrop's radius gives an increased value, which affects the new surface potential calculation, making it more negative because the same amount of charge is now spread out over a larger sphere. This results in the larger raindrop having a greater potential at its surface, which follows directly from the formula for calculating electric potential for spherical charge distributions.