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Consider a uranium nucleus to be sphere of radius \(R=7.4 \times 10^{-15} \mathrm{m}\) with a charge of \(92 e\) distributed uniformly throughout its volume. (a) What is the electric force exerted on an electron when it is \(3.0 \times 10^{-15} \mathrm{m}\) from the center of the nucleus? (b) What is the acceleration of the electron at this point?

Short Answer

Expert verified
The electric force exerted on the electron when it is \(3.0 \times 10^{-15} \mathrm{m}\) from the center of the uranium nucleus is approximately \(-3.15 \times 10^{-3} \mathrm{N}\), and the acceleration of the electron at this point is approximately \(-3.46 \times 10^{27} \mathrm{m/s^2}\).

Step by step solution

01

Calculating the Electric Field at 3.0 x 10^-15 m from the Center of the Nucleus

To calculate the electric field at the given distance from the center of the nucleus, we will be using the following formula derived from Gauss' law for a uniformly distributed spherical charge: \[E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}\] where \(E\) is the electric field, \(\varepsilon_0\) is the permittivity of free space, \(Q\) is the charge of the uranium nucleus, and \(r\) is the given distance from the center of the nucleus. The value for the permittivity of free space is: \[\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{C^2/N.m^2}\] The charge of the uranium nucleus is given as \(92 e\), where \(e\) is the elementary charge: \[e = 1.60 \times 10^{-19} \mathrm{C}\] Now plug in the given values and determine the electric field:
02

Calculate the Electric Field at the given distance

\[E = \frac{1}{4 \pi \varepsilon_0} \frac{92e}{(3.0 \times 10^{-15} \mathrm{m})^2}\] \[E = \frac{1}{4 \pi (8.85 \times 10^{-12} \mathrm{C^2/N.m^2})} \frac{92(1.60 \times 10^{-19} \mathrm{C})}{(3.0 \times 10^{-15} \mathrm{m})^2}\] After calculating the above expression, we find the electric field: \[E \approx 1.97 \times 10^{16} \mathrm{N/C}\]
03

Calculate the Electric Force exerted on the Electron

Now that we have the electric field at the given distance, we can find the electric force exerted on the electron. The force experienced by a charged particle in an electric field is given by the equation: \[F = qE\] where \(F\) is the force, \(q\) is the charge of the electron, and \(E\) is the electric field. The charge of the electron is: \[q = -e = -1.60 \times 10^{-19} \mathrm{C}\] Plug in the values and calculate the force: \[F = (-1.60 \times 10^{-19} \mathrm{C})(1.97 \times 10^{16} \mathrm{N/C})\] \[F \approx -3.15 \times 10^{-3} \mathrm{N}\] So the electric force exerted on the electron is approximately -3.15 x 10^-3 N.
04

Calculate the Acceleration of the Electron

For the final step, we need to find the acceleration of the electron. The acceleration is given by Newton's second law: \[F = ma\] where \(F\) is the force, \(m\) is the mass of the electron, and \(a\) is the acceleration. The mass of the electron is: \[m = 9.11 \times 10^{-31} \mathrm{kg}\] Solve for the acceleration: \[a = \frac{F}{m}\] \[a = \frac{-3.15 \times 10^{-3} \mathrm{N}}{9.11 \times 10^{-31} \mathrm{kg}}\] After calculating the above expression, we find the acceleration: \[a \approx -3.46 \times 10^{27} \mathrm{m/s^2}\] Thus, the acceleration of the electron at the given distance from the center of the uranium nucleus is approximately -3.46 x 10^27 m/s².

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Key Concepts

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

Gauss' Law
Gauss' Law is a fundamental principle that relates the distribution of electric charge to the resulting electric field. It is one of the four Maxwell's equations, which are the foundation of classical electromagnetism. Simply put, Gauss' Law states that the total electric flux through a closed surface is proportional to the enclosed electric charge. The law can be mathematically expressed as:

\[\begin{equation}\Phi_E = \oint{\mathbf{E} \cdot d\mathbf{A}} = \frac{Q_{\text{enc}}}{\varepsilon_0}\end{equation}\]
where \(\Phi_E\) is the electric flux, \(\mathbf{E}\) is the electric field, \(d\mathbf{A}\) represents an infinitesimal area on the closed surface, \(Q_{\text{enc}}\) is the enclosed charge, and \(\varepsilon_0\) is the permittivity of free space. In simpler terms, it provides a way to calculate the electric field created by symmetrical charge distributions, such as the spherical uranium nucleus in our exercise.
Electric Charge
Electric charge is a fundamental property of particles that causes them to experience a force when placed in an electric and magnetic field. Charges come in two types, commonly referred to as positive and negative. Like charges repel each other while opposite charges attract. The unit of charge is the Coulomb (\(C\)), with the elementary charge being approximately \(1.60 \times 10^{-19} C\). In our exercise, the uranium nucleus has a charge of \(92e\), or 92 times the elementary charge, amounting to a sizable positive charge that influences the behavior of nearby charged particles, like electrons.
Uniformly Charged Sphere
A uniformly charged sphere is an idealized model where charge is spread evenly across the volume of a sphere. This model simplifies calculations of the electric field because, by symmetry, the field at any point outside the sphere depends only on the total charge and the distance from the center of the sphere. For points inside the sphere, the electric field varies linearly with distance from the center. These properties are essential for solving problems like the one in the exercise, where we need to know the electric force a distance away from the center of a charged spherical nucleus.
Electric Force on an Electron
The electric force on an electron in an electric field is determined by Coulomb's law, which is directly related to the concept of an electric field. The force \(F\) on a charge \(q\) in an electric field \(E\) is given by the equation \(F = qE\). Since an electron has a negative charge, it experiences a force in the direction opposite to the electric field. In our exercise, this is demonstrated by the electron being repelled by the positively charged uranium nucleus. By knowing the magnitude of the electric field and the charge on the electron, we calculated the force exerted on the electron and its subsequent acceleration using Newton's second law. It's vital to remember that the electron will move in response to this force, revealing how charged particles behave in the presence of an electromagnetic field.

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Most popular questions from this chapter

A point charge \(q\) is located at the center of a cube whose sides are of length \(a\). If there are no other charges in this system, what is the electric flux through one face of the cube?

Two parallel conducting plates, each of cross-sectional area \(400 \mathrm{cm}^{2},\) are \(2.0 \mathrm{cm}\) apart and uncharged. If \(1.0 \times 10^{12}\) electrons are transferred from one plate to the other, what are (a) the charge density on each plate? (b) The electric field between the plates?

A total charge \(5.0 \times 10^{-6} \mathrm{C}\) is distributed uniformly throughout a cubical volume whose edges are \(8.0 \mathrm{cm}\) long. (a) What is the charge density in the cube? (b) What is the electric flux through a cube with 12.0 -cm edges that is concentric with the charge distribution? (c) Do the same calculation for cubes whose edges are \(10.0 \mathrm{cm}\) long and 5.0 cm long. (d) What is the electric flux through a spherical surface of radius \(3.0 \mathrm{cm}\) that is also concentric with the charge distribution?

An infinite charged wire with charge per unit length \(\lambda\) lies along the central axis of a cylindrical surface of radius \(r\) and length \(l .\) What is the flux through the surface due to the electric field of the charged wire?

A spherical rubber balloon carries a total charge \(Q\) distributed uniformly over its surface. At \(t=0,\) the radius of the balloon is \(R\). The balloon is then slowly inflated until its radius reaches \(2 R\) at the time \(t_{0} .\) Determine the electric field due to this charge as a function of time (a) at the surface of the balloon, (b) at the surface of radius \(R,\) and \((\mathrm{c})\) at the surface of radius \(2 R\). Ignore any effect on the electric field due to the material of the balloon and assume that the radius increases uniformly with time.

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