Problem 72 - University Physics with Modern Physics

Question

(a) If a spherical raindrop of radius 0.650 mm carries a charge of \(-\)3.60 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) -49.8 V; (b) Radius = 0.819 mm, Potential = -79.0 V.

Step 1: Calculate the Potential on the Surface of the Spherical Raindrop

The formula to calculate the potential due to a charged sphere at its surface is given by \[ V = \frac{kQ}{r} \]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.Plugging in the values: \( Q = -3.60 \, \text{pC} = -3.60 \times 10^{-12} \, \text{C} \) and \( r = 0.650 \, \text{mm} = 0.650 \times 10^{-3} \, \text{m} \), we have:\[ V = \frac{(8.99 \times 10^9)(-3.60 \times 10^{-12})}{0.650 \times 10^{-3}} \]Solving this gives:\[ V = -49.8 \, \text{V} \]

Step 2: Determine the Charge and Properties of the Larger Raindrop

When two identical spherical raindrops merge, the charge \( Q \) is additive:\[ Q_{\text{total}} = 2Q = 2(-3.60 \times 10^{-12} \text{ C}) = -7.20 \times 10^{-12} \text{ C} \]The volume of a sphere is given by:\[ V = \frac{4}{3} \pi r^3 \]Since the total volume after merging is twice the volume of a single drop:\[ 2V = \frac{4}{3} \pi (0.650 \times 10^{-3})^3 \times 2 \]The radius of the larger drop \( R \) is given by solving for the volume:\[R^3 = 2r^3 \R = 2^{1/3}r = 2^{1/3} \times 0.650 \times 10^{-3} \, \text{m} = 0.819 \times 10^{-3} \, \text{m}\]

Step 3: Calculate the Potential on the Surface of the Larger Raindrop

Now calculate the potential at the surface of the larger raindrop using the formula:\[ V_{\text{larger}} = \frac{kQ_{\text{total}}}{R} \]where \( Q_{\text{total}} = -7.20 \times 10^{-12} \, \text{C} \) and \( R = 0.819 \times 10^{-3} \, \text{m} \).Plug in the values:\[ V_{\text{larger}} = \frac{(8.99 \times 10^9)(-7.20 \times 10^{-12})}{0.819 \times 10^{-3}} \]Solving this gives:\[ V_{\text{larger}} = -79.0 \, \text{V} \]

Key Concepts

Coulomb's Law

Coulomb's Law is one of the fundamental principles of electrostatics, describing the force between two charged objects. According to this law, the electric force between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as: \[ F = k \frac{|q_1 q_2|}{r^2} \]where:

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

In our exercise, we didn't calculate force but needed to find the electric potential at the surface of a charged raindrop. However, understanding Coulomb's Law is essential as it underpins the calculations used to determine electric potential.

Electric Potential

Electric potential is a measure of the electric potential energy per unit charge at a specific point in space due to electric fields. The potential at a point indicates how much work is required to move a unit positive charge from a reference point, typically at infinity, to that point.For a charged sphere, like our spherical raindrop, the electric potential (\( V \)) at its surface can be calculated using:\[ V = \frac{kQ}{r} \]where:

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

In the exercise, we used this formula to find the potential at the surface of a spherical raindrop and then recalculated for a larger raindrop formed by merging two of these drops. Calculating electric potential helps us understand how charged surfaces interact with nearby objects and fields.

Sphere Merging

When two identical charged spheres, such as raindrops, merge, the resulting sphere is larger, and its properties must be recalculated. Here's how it works:

• **Charge**: The charge of the merged sphere is simply the sum of the charges on the individual spheres. In our case, the total charge \( Q_{\text{total}} \) is \(-7.20 \times 10^{-12} \, \text{C} \), double the charge of a single raindrop.
• **Volume**: The volume of the new sphere is twice the volume of one original raindrop. From this, we can derive the radius of the new sphere using the volume formula for a sphere \( V = \frac{4}{3} \pi r^3 \). Solving for the radius gives \( r^3 = 2r_0^3 \), linking the new radius to the original radius.
• **Potential**: With the new charge and radius, we recalculate the electric potential using the same formula \( V = \frac{kQ_{\text{total}}}{R} \). This shows how the merging of the two spheres affects the potential due to the increased size and charge distribution of the new raindrop.

Understanding sphere merging in electrostatics is crucial for applying concepts of charge conservation and calculating resulting field properties.