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Chapter 18: Problem 64

(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?

Short Answer

Expert verified
(a) -16.6 volts; (b) Radius: 0.819 mm, Potential: -26.3 volts.

Step by step solution

01

Calculate Potential of a Single Raindrop

The electric potential at the surface of a single charged sphere can be calculated using the formula \[ V = \frac{k_e \cdot Q}{R} \]where:- \( V \) is the electric potential,- \( k_e = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant,- \( Q = -1.20 \, \text{pC} = -1.20 \times 10^{-12} \, \text{C} \) is the charge,- \( R = 0.650 \, \text{mm} = 0.650 \times 10^{-3} \, \text{m} \) is the radius.Inserting the values, we get:\[ V = \frac{8.99 \times 10^9 \cdot (-1.20 \times 10^{-12})}{0.650 \times 10^{-3}} = -16.6 \, \text{volts} \]
02

Determine Radius of Merged Raindrop

When two identical spheres merge, their volumes combine. The original volume is given by \[ V_{\text{initial}} = \frac{4}{3} \pi R^3 \]For two raindrops, volume doubles:\[ 2V_{\text{initial}} = \frac{4}{3} \pi (0.650 \times 10^{-3})^3 \times 2 \]To find the new radius \( R_f \) of the merged drop, equate this to the final volume:\[ V_{\text{final}} = \frac{4}{3} \pi R_f^3 = \frac{8}{3} \pi R^3 \]Cancelling terms gives:\[ R_f^3 = 2R^3 \]\[ R_f = (2R^3)^{1/3} \]Substituting \( R = 0.650 \times 10^{-3} \) m:\[ R_f = 0.650 \times 10^{-3} \times (2)^{1/3} \approx 0.819 \times 10^{-3} \text{ } \text{m} \]
03

Calculate Potential of Merged Raindrop

Both raindrops have a combined charge since they merged, so total charge is\[ Q_{\text{total}} = 2 \times (-1.20 \times 10^{-12}) = -2.40 \times 10^{-12} \, \text{C} \]The potential at the surface of the larger raindrop is:\[ V_{\text{merged}} = \frac{k_e \cdot Q_{\text{total}}}{R_f} \]Substitute the values:\[ V_{\text{merged}} = \frac{8.99 \times 10^9 \cdot (-2.40 \times 10^{-12})}{0.819 \times 10^{-3}} \approx -26.3 \, \text{volts} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical Raindrop
A spherical raindrop, as the name suggests, is shaped like a sphere and is characterized by its radius. This simple geometric shape makes it easy to apply many equations from physics, especially when addressing charge and electrical potential. For a sphere, the surface is the boundary, and any charge it carries is uniformly distributed if stated in the problem. A sphere's volume is calculated by the formula \( V = \frac{4}{3} \pi R^3 \), where \( R \) is the radius. This formula is crucial when calculating changes in dimensions due to merging or splitting.
Coulomb's Constant
Coulomb's constant, \( k_e \), is a fundamental value in electrostatics. It is used in the calculation of electric forces and potentials between charged objects. The value of Coulomb's constant is approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \). With this constant, we can relate the force between two charges directly to their product and inversely to the square of their separation. In the given exercise, \( k_e \) helps compute the electric potential at the surface of raindrops by magnifying the effect of the charge over a certain distance.
Charge Distribution
Charge distribution refers to how charge is spread over a material or body. In the context of the spherical raindrop, the charge is uniformly distributed over the volume. This means each equal volume element of the raindrop contains the same amount of charge. Uniform distribution simplifies calculations, allowing use of the electric potential formula for spheres: \( V = \frac{k_e \cdot Q}{R} \). This approach assumes an even spread, which is vital for simplifying and solving problems in electrostatics effectively.
Merged Raindrop
A merged raindrop results from the collision and union of two or more droplets. With their combining, the total volume and total charge change. The volume of the merged raindrop is the sum of individual volumes, leading to a new radius calculated by \( R_f = (2R^3)^{1/3} \) when two identical droplets merge. The charge of the merged raindrop is simply the sum of the individual charges. Hence, in such scenarios, we calculate the new potential by recalculating using this combined charge and new radius, solidifying the link between geometry and electrical concepts.

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Most popular questions from this chapter

In the Bohr model of the hydrogen atom, a single electron revolves around a single proton in a circle of radius \(r .\) Assume that the proton remains at rest. (a) By equating the electric force to the electron mass times its acceleration, derive an expression for the electron's speed. (b) Obtain an expression for the electron's kinetic energy, and show that its magnitude is just half that of the electric potential energy. (c) Obtain an expression for the total energy, and evaluate it using \(r=5.29 \times 10^{-11} \mathrm{~m}\). Give your numerical result in joules and in electronvolts.

Two very large metal parallel plates that are \(25 \mathrm{~cm}\) apart, oriented perpendicular to a sheet of paper, are connected across the terminals of a \(50.0 \mathrm{~V}\) battery. (a) Draw to scale the lines where the equipotential surfaces due to these plates intersect the paper. limit your drawing to the region between the plates, avoiding their edges, and draw the lines for surfaces that are \(10.0 \mathrm{~V}\) apart, starting at the lowpotential plate. (b) These surfaces are separated equally in potential. Are they also separated equally in distance? (c) In words, describe the shape and orientation of the surfaces you just found.

The paper dielectric in a paper-and-foil capacitor is \(0.0800 \mathrm{~mm}\) thick. Its dielectric constant is \(2.50,\) and its dielectric strength is \(50.0 \mathrm{MV} / \mathrm{m} .\) Assume that the geometry is that of a parallel- plate capacitor, with the metal foil serving as the plates. (a) What area of each plate is required for a \(0.200 \mu \mathrm{F}\) capacitor? (b) If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the capacitor?

Two oppositely charged identical insulating spheres, each \(50.0 \mathrm{~cm}\) in diameter and carrying a uniform charge of magnitude \(175 \mu \mathrm{C},\) are placed \(1.00 \mathrm{~m}\) apart center to center (Figure 18.48 ). (a) If a voltmeter is connected between the nearest points \((a\) and \(b\) ) on their surfaces, what will it read? (b) Which point, \(a\) or \(b\), is at the higher potential? How can you know this without any calculations?

Two protons are released from rest when they are \(0.750 \mathrm{nm}\) apart. (a) What is the maximum speed they will reach? When does this speed occur? (b) What is the maximum acceleration they will achieve? When does this acceleration occur?
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