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Ch 37: The Foundations of Modern Physics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 37: The Foundations of Modern PhysicsProblema 11c
Capitolo 37, Problema 11c

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if the upper electrode is 20 V more positive than the lower electrode. The density of the oil is 885 kg/m3. Does the droplet have a surplus or a deficit of electrons? How many?

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1
Determine the force acting on the droplet due to gravity. The gravitational force is given by \( F_g = m g \), where \( m \) is the mass of the droplet and \( g \) is the acceleration due to gravity. To find \( m \), calculate the volume of the spherical droplet using \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the droplet (half the diameter). Then, use the density \( \rho \) to find the mass: \( m = \rho V \).
Determine the electric force acting on the droplet. Since the droplet is motionless, the electric force \( F_e \) must balance the gravitational force \( F_g \). The electric force is given by \( F_e = q E \), where \( q \) is the charge on the droplet and \( E \) is the electric field between the electrodes. The electric field can be calculated using \( E = \frac{\Delta V}{d} \), where \( \Delta V \) is the potential difference (20 V) and \( d \) is the separation between the electrodes (11 mm).
Set the magnitudes of the forces equal to each other to solve for the charge \( q \): \( F_g = F_e \), or \( m g = q E \). Rearrange to find \( q \): \( q = \frac{m g}{E} \). Substitute the expressions for \( m \) and \( E \) to calculate \( q \).
Determine whether the droplet has a surplus or deficit of electrons. Since the upper electrode is more positive, the electric field points downward. For the droplet to remain motionless, the electric force must point upward, meaning the droplet must have a negative charge (surplus of electrons).
Calculate the number of excess electrons on the droplet. The charge of a single electron is \( e = 1.6 \times 10^{-19} \; \text{C} \). The number of excess electrons \( n \) is given by \( n = \frac{q}{e} \). Use the value of \( q \) calculated in the previous step to find \( n \).

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Electric Force

The electric force is the interaction between charged particles, described by Coulomb's law. In this scenario, the electric field created by the voltage difference between the electrodes exerts a force on the charged oil droplet. This force can either balance the gravitational force acting on the droplet or cause it to move, depending on the charge of the droplet.
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Work due to Electric Force

Gravitational Force

The gravitational force is the attractive force between two masses, calculated using Newton's law of universal gravitation. For the oil droplet, this force is determined by its mass, which can be derived from its volume and density. The droplet's weight must be balanced by the electric force for it to remain motionless between the electrodes.
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Gravitational Forces in 2D

Charge of an Electron

The charge of an electron is a fundamental physical constant, approximately -1.6 x 10^-19 coulombs. To determine whether the droplet has a surplus or deficit of electrons, one must calculate the net charge required to balance the gravitational force with the electric force. The number of excess or deficit electrons can then be found by dividing the total charge by the charge of a single electron.
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Electrons In Water (Using Density)
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