Problem 80 - University Physics with Modern Physics

Question

(a) If a spherical raindrop of radius 0.650 \(\mathrm{mm}\) carries a charge of \(-1.20 \mathrm{pC}\) uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.) (b) Two identical raindrops, each with radius and charge specified in part (a), collide and merge into one larger raindrop. What is the radius of this larger drop, and what is the potential at its surface, if its charge is uniformly distributed over its volume?

Step-by-Step Solution

Answer

(a) The potential is \(-16.6 \, \text{V}\). (b) The radius is \(0.819 \, \text{mm}\) and the potential is \(-26.3 \, \text{V}\).

Step 1: Understand the formula for electric potential

The electric potential due to a charged sphere can be calculated using the formula: \ V = \frac{kQ}{r} \ where \(V\) is the potential, \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\), \(Q\) is the charge, and \(r\) is the radius of the sphere.

Step 2: Calculate potential for a single raindrop

Substitute \(Q = -1.20 \, \text{pC} = -1.20 \times 10^{-12} \, \text{C}\) and \(r = 0.650 \, \text{mm} = 0.650 \times 10^{-3} \, \text{m}\) into the formula: \ V = \frac{8.99 \times 10^9 \times (-1.20 \times 10^{-12})}{0.650 \times 10^{-3}} \ V \approx -16.6 \, \text{V}. \ Thus, the potential at the surface of a single raindrop is \(-16.6 \, \text{V}\).

Step 3: Determine the volume of the larger raindrop

When two raindrops merge, their volumes add. The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\). The volume of the larger drop \(V_{\text{larger}} = 2 \times \frac{4}{3} \pi (0.650 \times 10^{-3})^3 \, \text{m}^3\).

Step 4: Calculate radius of the larger drop

The volume of the merged drop is \(V_{\text{larger}} = \frac{4}{3} \pi R^3\). Equating the volumes: \ 2 \times \frac{4}{3} \pi (0.650 \times 10^{-3})^3 = \frac{4}{3} \pi R^3 \ Solve for \(R\): \ R^3 = 2 \times (0.650 \times 10^{-3})^3 \ R = (2 \times 0.650^3)^{1/3} \times 10^{-3} \ R \approx 0.819 \, ext{mm}. \ The radius of the merged drop is approximately \(0.819 \, \text{mm}\).

Step 5: Calculate potential at the surface of the larger drop

The total charge \(Q_{\text{total}}\) is \(-2.40 \, \text{pC}\) (sum of charges from both raindrops). The potential at the surface is: \ V = \frac{8.99 \times 10^9 \times (-2.40 \times 10^{-12})}{0.819 \times 10^{-3}} \ V \approx -26.3 \, \text{V}. \ Thus, the potential at the surface of the larger raindrop is \(-26.3 \, \text{V}\).

Key Concepts

Coulomb's Law

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

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.

Raindrop Collision

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.