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Chapter 2: Problem 28

A thin disk, radius \(3 \mathrm{~cm}\), has a circular hole of radius \(1 \mathrm{~cm}\) in the middle. There is a uniform surface charge of \(-4 \mathrm{esu} / \mathrm{cm}^{2}\) on the disk. (a) What is the potential in statvolts at the center of the hole? (Assume zero potential at infinite distance.) (b) An electron, starting from rest at the center of the hole, moves out along the axis, experiencing no forces except repulsion by the charges on the disk. What velocity does it ultimately attain? (Electron mass \(=9 \times 10^{-28} \mathrm{gm} .\) )

Short Answer

Expert verified
(a) The potential is \\(-24\pi\\) statvolts. (b) The electron's speed is \\3.097 \times 10^9\\ cm/s.

Step by step solution

01

Understand the Problem

We have a thin disk with a circular hole cut out in the center, and there is a uniform surface charge density on the disk. We will calculate the potential at the center of the hole first, and then determine the final velocity of an electron starting from the center.
02

Calculate Electric Potential at the Hole's Center

Since the potential is scalar, we can calculate it as a superposition of the potential due to the entire disk without the hole and the potential of the negative charge of the material removed from the hole. The disk's contribution is \(V_{disk} = -\frac{2 \pi \sigma (R-r)^2}{\epsilon_0}\) and the hole's is \(V_{hole} = \frac{2 \pi \sigma r^2}{\epsilon_0}\). Where \(\sigma = -4\ \mathrm{esu/cm^2}\). Thus, the potential at the center is \(V = V_{disk} - V_{hole}.\)
03

Evaluate \(V_{disk}\) and \(V_{hole}\)

For \(V_{disk}\): \(\sigma = -4\ \mathrm{esu/cm^2},\ R = 3\ \mathrm{cm},\ r = 1\ \mathrm{cm}.\) Since we're using constructor units, the constants simplify the expression, hence, \(V_{disk} = 2\pi\cdot(-4)\cdot 2^2 = -32\pi\). \(V_{hole} = 2\pi\cdot(-4)\cdot 1^2 = -8\pi\).
04

Compute the Total Potential at the Center of the Hole

The net potential at the center is given as \(V = V_{disk} - V_{hole} = -32\pi + 8\pi = -24\pi\).
05

Use Conservation of Energy to Find Electron's Velocity

The change in potential energy is equal to the change in kinetic energy. Since the electron starts from rest, its kinetic energy change is \(\Delta KE = eV = (4.8 \times 10^{-10}) (-24\pi) \). On equating with kinetic energy \(\frac{1}{2}mv^2\), with \(m = 9 \times 10^{-28}\mathrm{gm}\), solve for \(v\): \(v = \sqrt{\frac{2 \times 24\pi \times 4.8 \times 10^{-10}}{9 \times 10^{-28}}}\).
06

Calculate the Velocity

Simplifying the previous expression, we find \(v = \sqrt{\frac{84.48 \times 10^{-10}}{9 \times 10^{-28}}} = \sqrt{9.3867\times 10^{18}} = 3.097 \times 10^9\) cm/s.

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Key Concepts

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

Electric Potential
Electric potential is a fundamental concept in electrostatics. It refers to the work done in bringing a unit positive charge from infinity to a point in an electric field without any acceleration. This is often considered as the energy per unit charge. In simpler terms, electric potential can be understood as the electrical "height" at a point, similar to gravitational potential in mechanics.
  • A positive charge increases the electric potential, while a negative charge decreases it.
  • The potential is measured in volts in the standard international unit system.
  • It's important to note that electric potential is a scalar quantity, meaning it has magnitude but no direction.
In the given problem, the electric potential at the center of the hole is determined by considering both the disk and the hole. Since the potential contributes additively, we can use superposition principles to calculate the resultant potential at the designated point.
Surface Charge Density
Surface charge density is a measure of the amount of electric charge per unit area on a surface. It is denoted by the symbol \(\sigma\) and is generally measured in units such as coulombs per square meter or electrostatic units (esu) per square centimeter in the CGS unit system.
  • On a charged disk, surface charge density remains constant across its surface.
  • In the exercise, \(\sigma = -4 \ \,es\u \ material\s/cm^2\) is the uniform surface charge density of the disk.
  • The negative sign indicates the charge is negative, thus exerting a repulsive force on negatively charged particles like electrons.
This concept is crucial in determining effects like electric potential and force due to a charged surface, as it influences the entire electrostatic field generated by the surface.
Electron Velocity
Electron velocity is the speed at which an electron moves. In electrostatics, electrons follow the path of the electric field, moving from higher to lower potential areas. In this problem, the electron starts from rest, indicating it initially had no kinetic energy.
  • The velocity is derived from the conservation of energy, as potential energy changes into kinetic energy while the electron moves against the repulsive force.
  • The calculation involves solving \[ eV = \frac{1}{2}mv^2 \], where \( e \) is the electron charge and \( m \) its mass.
  • The unit of velocity is centimeters per second (cm/s) in this context.
Understanding electron velocity is important in predicting how electrons will behave when subjected to electric fields, which is essential in everything from electronics to particle physics.
Coulomb's Law
Coulomb鈥檚 Law describes the force between two point charges. According to the law, the magnitude of the electrostatic force between two charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between their centers. Mathematically, it is expressed as:\[ F = k_e \frac{|q_1q_2|}{r^2} \]
  • Here, \( k_e \) is the Coulomb constant; \( q_1 \) and \( q_2 \) are the charges; and \( r \) is the distance between them.
  • Coulomb's Law helps in understanding the electric forces and fields created by charged objects.
  • It states that like charges repel and opposite charges attract, which is foundational in electrostatics.
While Coulomb's Law is not explicitly solved in this exercise, it underpins much of the work in analyzing the effects of the charged disk on the electron, as these forces dictate the electron鈥檚 behavior.

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Most popular questions from this chapter

Consider a charge distribution which has the constant density \(\rho\) everywhere inside a cube of edge \(b\) and is zero everywhere outside that cube. Letting the electric potential \(\phi\) be zero at infinite distance from the cube of charge, denote by \(\phi_{0}\) the potential at the center of the cube and \(\phi_{1}\) the potential at a corner of the cube. Determine the ratio \(\phi_{0} / \phi_{1}\). The answer can be found with very little calculation by combining a dimensional argument with superposition. (Think about the potential at the center of a cube with the same charge density and with twice the edge length.)

To show that it takes more than direction and magnitude to make a vector, let's try to define a vector which we'll name squrl \(\mathbf{F}\) by a relation like Eq. 61 with the right-hand side squared: $$ (\operatorname{squrl} \mathbf{F}) \cdot \hat{\mathbf{i}}=\left[\lim _{a_{i} \rightarrow 0} \frac{\int_{C_{i}} \mathbf{F} \cdot d \mathbf{s}}{a_{i}}\right]^{2} $$ Prove that this does not define a vector. (Hint: Consider reversing the direction of f.).

A flat nonconducting sheet lies in the \(x y\) plane. The only charges in the system are on this sheet. In the half-space above the sheet, \(z>0\), the potential is \(\phi=\phi_{0} e^{-k z} \cos k x\), where \(\phi_{0}\) and \(k\) are constants. (a) Verify that \(\phi\) satisfies Laplace's equation in the space above the sheet. (b) What do the electric field lines look like? ( \(c\) ) Describe the charge distribution on the sheet.

Does the function \(f(x, y)=x^{2}+y^{2}\) satisfy the two-dimensional Laplace's equation? Does the function \(g(x, y)=x^{2}-y^{2}\) ? Sketch the latter function, calculate the gradient at the points \((x=\) \(0, y=1) ;(x=1, y=0) ;(x=0, y=-1) ;\) and \((x=-1, y=\) 0) and indicate by little arrows how these gradient vectors point.

Describe the electric field and the charge distribution that go with the following potential: $$ \begin{array}{ll} \phi=x^{2}+y^{2}+z^{2} \quad \text { for } x^{2}+y^{2}+z^{2}
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