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Chapter 18: Problem 34

In a particular Millikan oil-drop apparatus, the plates are 2.25 \(\mathrm{cm}\) apart. The oil used has a density of \(0.820 \mathrm{g} / \mathrm{cm}^{3},\) and the atomizer that sprays the oil drops produces drops of diameter \(1.00 \times 10^{-3} \mathrm{mm}\) . (a) What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains five excess electrons? (b) What should be the potential difference across the plates to produce this electric field? (c) If another drop of the same oil requires a plate potential of 73.8 \(\mathrm{V}\) to hold it stationary, how many excess electrons did it contain?

Short Answer

Expert verified
(a) 5250 N/C, (b) 118.13 V, (c) 8 electrons

Step by step solution

01

Calculate the Volume of the Oil Drop

First, find the volume of a spherical oil drop using the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] The diameter is given as \( 1.00 \times 10^{-3} \text{ mm} \), which is \( 1.00 \times 10^{-6} \text{ m} \). The radius \( r \) is half the diameter: \( r = 0.5 \times 10^{-6} \text{ m} \). Plugging into the formula, we get:\[ V = \frac{4}{3} \pi (0.5 \times 10^{-6})^3 \approx 5.24 \times 10^{-19} \text{ m}^3 \]
02

Determine the Mass of the Oil Drop

Next, calculate the mass using the equation \( m = \rho V \), where \( \rho \) is the density of the oil: \( 0.820 \text{ g/cm}^3 = 820 \text{ kg/m}^3 \). Thus, the mass \( m \) of the drop is:\[ m = 820 \times 5.24 \times 10^{-19} \approx 4.29 \times 10^{-16} \text{ kg} \]
03

Calculate the Gravitational Force on the Drop

The gravitational force \( F_g \) can be calculated using \( F_g = mg \), where \( g = 9.8 \text{ m/s}^2 \). Thus:\[ F_g = 4.29 \times 10^{-16} \times 9.8 \approx 4.20 \times 10^{-15} \text{ N} \]
04

Determine the Electric Force Required

For the drop to stay stationary, the electric force \( F_e \) must balance the gravitational force \( F_g \). Therefore, \( F_e = F_g \). The electric force is given by \( F_e = Eq \), where \( q = ne \) is the charge of the drop with \( n = 5 \) excess electrons and \( e = 1.6 \times 10^{-19} \text{ C} \). So, \( q = 5 \times 1.6 \times 10^{-19} \text{ C} = 8 \times 10^{-19} \text{ C} \). Thus, setting \( F_e = 4.20 \times 10^{-15} \text{ N} \), we find:\[ E = \frac{F_e}{q} = \frac{4.20 \times 10^{-15}}{8 \times 10^{-19}} = 5.25 \times 10^3 \text{ N/C} \]
05

Calculate the Potential Difference Across the Plates

The potential difference \( V \) required to create an electric field \( E \) across plates separated by distance \( d \) is given by: \[ V = Ed \] The separation \( d = 2.25 \text{ cm} = 0.0225 \text{ m} \). Thus: \[ V = 5.25 \times 10^3 \times 0.0225 = 118.13 \text{ V} \]
06

Calculate the Number of Excess Electrons for Another Drop

For another drop requiring a potential of \( 73.8 \text{ V} \):Use \( V = Ed \) to find \( E \):\[ E = \frac{V}{d} = \frac{73.8}{0.0225} = 3.28 \times 10^3 \text{ N/C} \] Now, find \( q \) using \( F_e = Eq = F_g = 4.20 \times 10^{-15} \text{ N} \):\[ q = \frac{4.20 \times 10^{-15}}{3.28 \times 10^3} = 1.28 \times 10^{-18} \text{ C} \] The number of excess electrons \( n \) is:\[ n = \frac{q}{e} = \frac{1.28 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 8 \]

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Key Concepts

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

Electric Field
An electric field is a region around a charged object that influences other charged objects within that space. In physics, the strength of the electric field is described by how much force a charge experiences when placed in the field. It is measured in newtons per coulomb (N/C).

In the Millikan experiment scenario, the electric field was responsible for balancing the gravitational force on the oil drop. The electric field exerts a force on any charges within it, calculated by the formula:
  • Electric Force (\( F_e \)) = Electric Field (\( E \)) 脳 Charge (\( q \))
The fundamental concept here is that the electric field must be strong enough to exert an upward force equal to the downward gravitational force on the oil droplet, thus keeping it stationary.
Potential Difference
Potential difference, often referred to as voltage, is the work needed to move a unit charge between two points in an electric field. It is measured in volts (V).

In the Millikan oil-drop experiment, the potential difference between two parallel plates creates the electric field that can hold a charged oil droplet stationary. The relationship between potential difference and the electric field across two plates separated by distance (\( d \)) is given by:
  • Potential Difference (\( V \)) = Electric Field (\( E \)) 脳 Distance (\( d \))
This equation shows how a greater potential difference is needed for a stronger electric field to maintain the balance of forces on the droplet.
Excess Electrons
Excess electrons refer to the extra electrons that an object, like an oil drop in the Millikan experiment, carries. Electrons are negatively charged particles, and having excess ones gives the droplet a net negative charge.

The number of excess electrons affects the total charge of the droplet, which is calculated by:
  • Total Charge (\( q \)) = Number of Excess Electrons (\( n \)) 脳 Elementary Charge (\( e \))
In the Millikan experiment, this charge is pivotal because it determines how strongly the droplet is influenced by the electric field and therefore the strength of the field required to counteract gravity.
Density
Density measures the mass per unit volume of a substance and is expressed in grams per cubic centimeter (g/cm鲁) or kilograms per cubic meter (kg/m鲁).

For the oil drop in the Millikan experiment, density is an essential factor since it dictates the mass of the oil drop, calculated as:
  • Mass (\( m \)) = Density (\( \rho \)) 脳 Volume (\( V \))
The mass then affects the gravitational force acting on the droplet. Understanding density is crucial to calculating how much force is necessary to counteract gravity and keep the droplet stationary with the electric field.
Gravitational Force
Gravitational force is the force exerted by the Earth on objects towards its center. It is calculated by the formula:
  • Gravitational Force (\( F_g \)) = Mass (\( m \)) 脳 Gravitational Acceleration (\( g \))
Here, gravitational acceleration is considered as 9.8 m/s虏. In the context of the Millikan experiment, the gravitational force pulls the oil drop downward. The goal of the experiment is to set the electric force equal to this gravitational force, allowing the droplet to hover in place without falling or rising. Understanding how gravitational force works is integral to determining the required electric field to achieve equilibrium.

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Most popular questions from this chapter

BIO The electric egg. The eggs of many species undergo a rapid change in the electrical potential difference across the outer membrane when they are fertilized. This change in potential difference affects the physiological development of the eggs. The poterntial difference across the membrane is called the membrane potential, \(V_{m},\) defined as the inside potential minus the outside potential. The membrane potential \(V_{m}\) arises when protein enzymes use the energy available in ATP to actively expel sodium ions (Na') and accumulate potassium ions \(\left(\mathrm{K}^{+}\right) .\) Because the membrane of the unfertilized egg is selectively permeable to \(\mathrm{K}^{+},\) the \(V_{m}\) of the resting sea urchin egg is about \(-70 \mathrm{mV}\) ; that is, the inside has a potential of 70 \(\mathrm{mV}\) less than that of the outside. The egg membrane behaves as a capacitor with a specific capacitance of about 1\(\mu \mathrm{F} / \mathrm{cm}^{2} .\) When a sea urchin egg is fertilized, Na' channels in the membrane are opened, \(\mathrm{Na}^{+}\) enters the egg, and \(V_{m}\) rapidly changes to \(+30 \mathrm{mV},\) where it remains for several minutes. The concentration of \(\mathrm{Na}^{+}\) in the egg's interior is about 30 mmoles/liter (30 \(\mathrm{mM} )\) and 450 \(\mathrm{mM}\) in the surrounding sea water. The inside \(\mathrm{K}^{*}\) concentration is about 200 \(\mathrm{mM}\) and the outside \(\mathrm{K}^{+}\) is 10 \(\mathrm{mM} .\) A useful constant that connects electrical and chemical units is the Faraday number, which has a value of approximately \(10^{5}\) coulomb/mole. That is, an Avogadro number (a mole) of monovalent ions such as Na^ + or \(\mathrm{K}^{+}\) carries a charge of \(10^{5} \mathrm{C}\) . How many moles of \(\mathrm{Na}^{+}\) must move per unit area of membrane to change \(V_{m}\) from \(-70 \mathrm{mV}\) to \(+30 \mathrm{mV},\) making the assumption that the membrane behaves purely as a capacitor? A. \(10^{-4}\) mole \(/ \mathrm{cm}^{2}\) B. \(10^{-9}\) mole/cm \(^{2}\) C. \(10^{-12} \mathrm{mole} / \mathrm{cm}^{2}\) D. \(10^{-14} \mathrm{mole} / \mathrm{cm}^{2}\)

The paper dielectric in a paper-and-foil capacitor is 0.0800 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\(\mu F\) 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?

A proton and an alpha particle are released from rest when they are 0.225 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur, just following the release of the particles or after a very long time?

(a) How much charge does a battery have to supply to a 5.0\(\mu \mathrm{F}\) capacitor to create a potential difference of 1.5 \(\mathrm{V}\) across its plates? How much energy is stored in the capacitor in this case? (b) How much charge would the battery have to supply to store 1.0 \(\mathrm{J}\) of energy in the capacitor? What would be the potential across the capacitor in that case?

A charge of 28.0 \(\mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and that has a magnitude of \(4.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) . What work is done by the electric force when the charge moves (a) 0.450 \(\mathrm{m}\) to the right; (b) 0.670 \(\mathrm{m}\) upward; (c) 2.60 \(\mathrm{m}\) at an angle of \(45.0^{\circ}\) downward from the horizontal?
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