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The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately \(7.4 \times 10^{-15} \mathrm{m}\) (a) What is the electric field this nucleus produces just outside its surface? (b) What magnitude of electric field does it produce at the distance of the electrons, which is about \(1.0 \times 10^{-10} \mathrm{m} ?\) (c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?

Short Answer

Expert verified
(a) \(2.41 \times 10^{21} \text{ N/C}\); (b) \(1.32 \times 10^{12} \text{ N/C}\); (c) zero electric field.

Step by step solution

01

Understanding the Electric Field Formula

The electric field just outside a uniformly charged sphere can be calculated using Gauss's Law. For a sphere of charge with total charge \(Q\), the electric field \(E\) at a distance \(r\) from the center of the sphere is given by \[E = \frac{kQ}{r^2},\] where \(k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\) is Coulomb's constant.
02

Calculate Total Charge of the Uranium Nucleus

The uranium nucleus, with 92 protons, has a total charge \(Q\) given by \(Q = 92e\), where \(e = 1.6 \times 10^{-19} \text{ C}\) is the elementary charge. Thus, \(Q = 92 \times 1.6 \times 10^{-19} = 1.472 \times 10^{-17} \text{ C}.\)
03

Calculate Electric Field Just Outside the Nucleus

Using the electric field formula and substituting the values, just outside the surface of the nucleus (radius \(r = 7.4 \times 10^{-15} \text{ m}\)), we get:\[E = \frac{8.99 \times 10^9 \times 1.472 \times 10^{-17}}{(7.4 \times 10^{-15})^2} \approx 2.41 \times 10^{21} \text{ N/C}.\]
04

Calculate Electric Field at the Electron Distance

Substitute \(r = 1.0 \times 10^{-10} \text{ m}\) into the electric field formula:\[E = \frac{8.99 \times 10^9 \times 1.472 \times 10^{-17}}{(1.0 \times 10^{-10})^2} \approx 1.32 \times 10^{12} \text{ N/C}.\]
05

Understanding the Electric Field by the Electron Shell at the Nucleus

According to the shell theorem, a uniformly charged shell exerts no net electric field inside the shell. Thus, at the location of the nucleus inside an electron shell, the net electric field from the electrons is zero.

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle in electromagnetism that relates the electric fields and the distribution of electric charge in space. It provides a method to calculate electric fields of symmetrical charge distributions, like the charge inside a sphere. Under this law, the electric flux through a closed surface is directly proportional to the total charge enclosed within it.

For our specific problem involving a uranium nucleus, modeled as a spherically symmetric charged sphere, Gauss's Law helps us calculate the electric field just outside its surface.
  • The formula for the electric field using Gauss's Law is \(E = \frac{kQ}{r^2}\), where \(k\) is Coulomb's constant.
  • \(Q\) represents the total charge of the nucleus, which in this case with 92 protons, equates to \(92e\).
This simplicity comes from the spherical symmetry, making it easier to apply Gauss's Law as each point at a surface equidistant from the center should have the same electric field magnitude.
Coulomb's Law
Coulomb's Law is the foundation of our understanding of electric force and fields. It describes how the electric force between two charges depends on the distance and magnitude of those charges. According to Coulomb’s Law, the electric force \(F\) between two point charges \(q_1\) and \(q_2\) is given by
  • \(F = \frac{k \cdot |q_1 \cdot q_2|}{r^2},\) where \(k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\).
  • The force is attractive if the charges are opposite and repulsive if they are the same.
When we use Coulomb's Law in defining the electric field from continuous charge distributions like a uranium nucleus, it's all about integrating those point interactions.

In the problem, after finding the total charge \(Q = 1.472 \times 10^{-17} \, \text{C}\) for the uranium nucleus, Coulomb's Law is applied through the formula derived from Gauss's Law to get the electric field across different points.
Nuclear Charge
Understanding the concept of nuclear charge is essential when dealing with atoms, particularly large ones like uranium, which houses numerous protons in its nucleus. The nuclear charge is fundamentally the total positive charge from protons within the nucleus. In simple terms, it is calculated as:
  • Nuclear charge \(Q = Z \, (e)\), where \(Z\) is the atomic number and \(e\) is the elementary charge (\(1.6 \times 10^{-19} \, \, \text{C}\)).
  • In uranium's case, \(Z = 92\).
This charge creates a strong electric field just outside the nucleus. The electric field has significant implications for both the structure of the atom and interactions with other charges.

In our exercise, the nuclear charge helps determine the electric field at differing distances, like directly on the surface of the nucleus and further out at electron distances. The interactions of these fields guide electron formation and behavior, representing why nuclear charge is fundamental not only in electric field calculations but in understanding atomic structure as a whole.

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Most popular questions from this chapter

A hollow, conducting sphere with an outer radius of 0.250 \(\mathrm{m}\) and an inner radius of 0.200 \(\mathrm{m}\) has a uniform surface charge density of \(+6.37 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) A charge of \(-0.500 \mu \mathrm{C}\) is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

Thomson's Model of the Atom. In the early years of the 20 th century, a leading model of the structure of the atom was that of the English physicist J. J. Thomson (the discoverer of the electron). In Thomson's model, an atom consisted of a sphere of positively charged material in which were embedded negatively charged electrons, like chocolate chips in a ball of cookie dough. Consider such an atom consisting of one electron with mass \(m\) and charge \(-e,\) which may be regarded as a point charge, and a uniformly charged sphere of charge \(+e\) and radius \(R\) (a) Explain why the equilibrium position of the electron is at the center of the nucleus. (b) In Thomson's model, it was assumed that the positive material provided little or no resistance to the motion of the electron. If the electron is displaced from equilibrium by a distance less than \(R,\) show that the resulting motion of the electron will be simple harmonic, and calculate the frequency of oscillation. Hint: Review the definition of simple harmonic motion in Section \(14.2 .\) If it can be shown that the net force on the electron is of this form, then it follows that the motion is simple harmonic. Conversely, if the net force on the electron does not follow this form, the motion is not simple harmonic.) (c) By Thomson's time, it was known that excited atoms emit light waves of only certain frequencies. In his model, the frequency of emitted light is the same as the oscillation frequency of the electron or electrons in the atom. What would the radius of a Thomson-model atom have to be for it to produce red light of frequency \(4.57 \times 10^{14}\) Hz? Compare your answer to the radii of real atoms, which are of the order of \(10^{-10} \mathrm{m}\) (see Appendix For data about the electron). (d) If the electron were displaced from equilibrium by a distance greater than \(R,\) would the electron oscillate? Would its motion be simple harmonic? Explain your reasoning. (Historical note: In \(1910,\) the atomic nucleus was discovered, proving the Thomson model to be incorrect. An atom's positive charge is not spread over its volume as Thomson supposed, but is concentrated in the tiny nucleus of radius \(10^{-14}\) to \(10^{-15} \mathrm{m} . )\)

A point charge of \(-2.00 \mu C\) is located in the center of a spherical cavity of radius 6.50 \(\mathrm{cm}\) inside an insulating charged solid. The charge density in the solid is \(\rho=7.35 \times 10^{-4} \mathrm{C} / \mathrm{m}^{3}\) Calculate the electric field inside the solid at a distance of 9.50 \(\mathrm{cm}\) from the center of the cavity.

A small conducting spherical shell with inner radius \(a\) and outer radius \(b\) is concentric with a larger conducting spherical shell with inner radius \(c\) and outer radius \(d\) (Fig. P22.47). The inner shell has total charge \(+2 q,\) and the outer shell has charge \(+4 q .\) (a) Calculate the electric field (magnitude and direction) in terms of \(q\) and the distance \(r\) from the common center of the two shells for (i) \(r < a ;\) (ii) \(a < r < b ;\) (iii) \(b < r < c ;\) (iv) \(c < r < d\) ; (v) \(r > d .\) Show your results in a graph of the radial component of \(\vec{\boldsymbol{E}}\) as a function of \(r .\) (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (ii) inner surface of the large shell; (iv) outer surface of the large shell?

A negative charge \(-Q\) is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. (a) Is there any excess charge induced on the inner surface of the piece of metal? If so, find its sign and magnitude. (b) Is there any excess charge on the outside of the piece of metal? Why or why not? (c) Is there an electric field in the cavity? Explain. (d) Is there an electric field within the metal? Why or why not? Is there an electric field outside the piece of metal? Explain why or why not. (e) Would someone outside the solid measure an electric field due to the charge \(-Q ?\) Is it reasonable to say that the grounded conductor has shielded the region from the effects of the charge \(-Q ?\) In principle, could the same thing be done for gravity? Why or why not?

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