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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\(^{-15}\) 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\(^{-10}\) 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) 2.4 \(\times\) 10^{21} N/C, (b) 1.3 \(\times\) 10^{11} N/C, (c) 0 N/C.

Step by step solution

01

Identify the Charged Sphere Formula

For the electric field outside a spherically symmetric charge distribution, we use Gauss's Law. The electric field just outside a uniformly charged sphere with total charge \(Q\) and radius \(R\) is given by \[ E = \frac{kQ}{R^2}, \]where \(k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\) is Coulomb's constant.
02

Calculate Charge of Uranium Nucleus

The charge \(Q\) on the uranium nucleus with 92 protons is determined by multiplying the number of protons by the elementary charge \(e = 1.6 \times 10^{-19} \, \text{C}\) per proton. \[ Q = 92 \cdot 1.6 \times 10^{-19} \, \text{C} = 1.472 \times 10^{-17} \, \text{C}. \]
03

Compute Electric Field just outside the surface

Now, substitute \(Q\) and \(R = 7.4 \times 10^{-15} \, \text{m}\) into the formula for the electric field:\[ E = \frac{8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \cdot 1.472 \times 10^{-17} \, \text{C}}{(7.4 \times 10^{-15} \, \text{m})^2} = 2.4 \times 10^{21} \, \text{N/C}. \]
04

Calculate Electric Field at Electron Distance

Use the same formula, but at a new distance of \(r = 1.0 \times 10^{-10} \, \text{m}\):\[ E = \frac{8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 \cdot 1.472 \times 10^{-17} \, \text{C}}{(1.0 \times 10^{-10} \, \text{m})^2} = 1.3 \times 10^{11} \, \text{N/C}. \]
05

Determine Net Electric Field Due to Electron Shell

By Gauss's Law, the net electric field inside a uniformly charged spherical shell is zero. Therefore, at the location of the nucleus, within the electron shell, the electric field produced by 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 powerful tool for calculating electric fields, especially when dealing with symmetrical charge distributions like spheres. It states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, Gauss's Law is expressed as: \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \(\Phi_E\) is the electric flux, \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) is the differential area vector on the surface, and \(Q_{enc}\) is the enclosed charge. \(\varepsilon_0\) is the permittivity of free space.
For spherically symmetric charge distributions, such as a nucleus or a charged shell, this law simplifies the calculation of electric fields by utilizing the symmetry to determine \(\mathbf{E}\) at a point outside the sphere. This makes the math straightforward when calculating fields outside charged spheres.
Coulomb's Law
Coulomb's Law describes the fundamental interaction between point charges, laying the groundwork for understanding electric forces and fields. It states that the magnitude of the electric force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance separating them. The law is expressed as: \[ F = k \frac{|q_1 q_2|}{r^2} \]where \(F\) is the force, \(q_1\) and \(q_2\) are the charges, \(r\) is the distance between the charges, and \(k\) is Coulomb's constant \(k = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\).
While Coulomb's Law applies to point charges, its principles extend to distributed charges by considering each point's contribution to the overall field. This law is integral in calculations involving charged spheres, as it helps derive expressions for electric fields via Gauss's Law.
Charged Spheres
Charged spheres are a common scenario in electromagnetism problems because their symmetry simplifies calculations of electric fields. When dealing with charged spheres, it's crucial to distinguish between cases where we're looking at points inside the sphere versus outside the sphere.
  • Outside the Sphere: The sphere behaves like a point charge located at its center, simplifying calculations. Gauss's Law can be easily applied here to find the electric field.
  • Inside the Sphere (Solid Sphere): If the sphere is uniformly charged, the electric field inside varies linearly with the distance from the center due to the gradual inclusion of more charge.
  • Spherical Shell: When considering the electric field in regions inside a hollow spherical shell, Gauss's Law suggests the field is zero because no net charge is enclosed within a Gaussian surface placed inside.
Understanding these cases helps accurately assess how electric fields behave in complex systems.
Electric Field of a Nucleus
The electric field surrounding a nucleus is vital for understanding atomic structure and interactions. For a nucleus like that of uranium, which contains 92 protons, the nuclear charge significantly influences the electric field experienced by nearby electrons. The electric field just outside the nucleus can be calculated using the formula derived from Gauss’s Law: \[ E = \frac{kQ}{R^2} \]where \(Q\) is the total charge from the protons in the nucleus and \(R\) is the radius of the nucleus. The radius and the number of protons make the field very strong and concentrated near the nucleus.
At larger distances, such as at the orbit of the electrons, the field diminishes significantly. This decrease is due to the increasing distance as evidenced by the inverse square law. Additionally, if electrons surround the nucleus evenly, the shell principle of Gauss's Law indicates they form a net zero electric field at the nucleus center. This critical balance helps stabilize atomic structures.

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Most popular questions from this chapter

A very long uniform line of charge has charge per unit length 4.80 \(\mu\)C/m and lies along the \(x\)-axis. A second long uniform line of charge has charge per unit length -2.40 \(\mu\)C/m and is parallel to the x-axis at \(y =\) 0.400 m. What is the net electric field (magnitude and direction) at the following points on the \(y\)-axis: (a) \(y =\) 0.200 m and (b) \(y =\) 0.600 m?

A square insulating sheet 80.0 cm on a side is held horizontally. The sheet has 4.50 nC of charge spread uniformly over its area. (a) Calculate the electric field at a point 0.100 mm above the center of the sheet. (b) Estimate the electric field at a point 100 m above the center of the sheet. (c) Would the answers to parts (a) and (b) be different if the sheet were made of a conducting material? Why or why not?

A point charge \(q_1 =\) 4.00 nC is located on the \(x\)-axis at \(x =\) 2.00 m, and a second point charge \(q_2 = -\)6.00 nC is on the \(y\)-axis at \(y =\) 1.00 m. What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 12.0 cm, giving it a charge of -49.0 \(\mu\)C. Find the electric field (a) just inside the paint layer; (b) just outside the paint layer; (c) 5.00 cm outside the surface of the paint layer.

A hollow, conducting sphere with an outer radius of 0.250 m and an inner radius of 0.200 m has a uniform surface charge density of +6.37 \(\times\) 10\(^{-6}\) C/m\(^2\). A charge of -0.500 \(\mu\)C is now introduced at the center of 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?

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