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Chapter 1: Problem 16

Gauss's law and two point charges ** (a) Two point charges \(q\) are located at positions \(x=\pm \ell\). At points close to the origin on the \(x\) axis, find \(E_{x}\). At points close to the origin on the \(y\) axis, find \(E_{y}\). Make suitable approximations with \(x \ll \ell\) and \(y \ll \ell\) (b) Consider a small cylinder centered at the origin, with its axis along the \(x\) axis. The radius is \(r_{0}\) and the length is \(2 x_{0}\). Using your results from part (a), verify that there is zero flux through the cylinder, as required by Gauss's law.

Short Answer

Expert verified
The electric field at points close to the origin on the \(x\) axis is \(E_{x} = 0\). At points close to the origin on the \(y\) axis, the electric field is \(E_{y} = 2\frac{kqy}{\ell^3}\). The flux through a small cylinder centered at the origin is zero, agreeing with Gauss's law.

Step by step solution

01

Understanding the electric field of a point charge

We can find the electric field \(E\) a distance \(x\) away from a point charge \(q\) using the equation: \(E = \frac{kq}{x^2}\), where \(k\) is Coulomb's constant. Since there are two charges and they are symmetrically located, the net electric field at any point on the \(x\) axis is the sum of both fields.
02

Calculating \(E_{x}\)

Due to symmetry, electric fields produced by each point charge at the origin will be in opposite directions along the \(x\)-axis, hence they will cancel each other out and \(E_{x} = 0\). This is true even for points close to the origin on the \(x\) axis.
03

Calculating \(E_{y}\)

For points close to the origin on the \(y\) axis, we have to approximate and take into account that \(y \ll \ell\), hence the distance to the charges is approximately \(\ell\). The electric field components \(E_{y1}\) and \(E_{y2}\) created by charges at locations \(x=+\ell\) and \(x=-\ell\) respectively are: \(E_{y1} = E_{y2} = \frac{kqy}{(\ell^2+y^2)^{3/2}}\). Since \(y \ll \ell\), we can simplify this to: \(E_{y1} = E_{y2} = \frac{kqy}{\ell^3}\). Due to symmetry, the total electric field \(E_y\) on the y-axis is the sum of both field components: \(E_{y} = E_{y1} + E_{y2} = 2\frac{kqy}{\ell^3}\).
04

Verifying Gauss's law

From Gauss's law, the electric flux through a closed surface is zero if there is no net enclosed charge. We have to confirm this using the electric field we have computed. We consider a small cylinder centered at the origin, with its axis along the \(x\) axis. The radius is \(r_{0}\) and the length is \(2 x_{0}\). As found in step 2, there is no electric field along the \(x\)-axis because fields from both charges cancel out. Thus, there's no flux through the flat caps of the cylinder. Since there's no charge enclosed by this cylindrical surface, the electric field along it is zero as well, so there is no flux either. So, the net flux through the cylinder is zero, thereby agreeing with Gauss's law.

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Key Concepts

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

Electric Field
The electric field is a vector field representing the force exerted by a charge on a unit positive charge at any given point in space. It is essential to understand its behavior around point charges. The formula to calculate the electric field produced by a point charge is given by \(E = \frac{kq}{r^2}\), where \(k\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge. This field has both magnitude and direction. It points radially outward from positive charges and radially inward towards negative charges.
Because electric fields are vectors, when multiple charges are present, we need to consider both the magnitudes and the directions. The overall field at a point due to multiple charges is the vector sum of the fields created by each charge. This concept of superposition is key to understanding field interactions in more complex setups like our current exercise.
Point Charges
Point charges are idealized charges that are assumed to be concentrated at a single point in space. In many physics problems, they simplify the analysis because they allow us to use symmetry and mathematical simplicity.
In the given exercise, we have two point charges located symmetrically at \(x = +\ell\) and \(x = -\ell\). Such configuration allows us to make approximations due to symmetry. Symmetry ensures that certain components of the electric field might cancel each other out, especially when the measurement point is located on the symmetric axis. This property simplifies the calculations significantly and is a powerful tool in electrostatics.
Electric Flux
Electric flux is a measure of the number of electric field lines passing through a surface. It provides a way to quantify the strength and extent of an electric field across a given area. The formula for electric flux \(\Phi\) through a surface is given by \(\Phi = \int \mathbf{E} \cdot d\mathbf{A}\), where \(\mathbf{E}\) is the electric field and \(d\mathbf{A}\) is the differential area vector.
According to Gauss's law, the net electric flux through a closed surface is proportional to the charge enclosed within that surface. In our exercise, a cylinder centered at the origin has no net charge inside, leading to a net flux of zero. This agrees with Gauss's law, confirming that our electric field calculations align with fundamental physical principles.
Symmetry in Electric Fields
Symmetry plays a crucial role in solving electric field problems, especially in cases involving multiple charges or particular field configurations. When charges are symmetrically placed, like the point charges in our exercise, certain components of the electric field can cancel each other. This happens because the fields from the charges mirror each other across the plane of symmetry.
In practical terms, this symmetry means that on the \(x\)-axis, the electric fields cancel entirely due to their opposite directions, resulting in zero net field there. On the \(y\)-axis, the fields add constructively, simplifying the calculation to a single sum. Recognizing and utilizing symmetry reduces complex problems to more manageable ones, proving vital in both learning and applying electrostatics concepts.

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Most popular questions from this chapter

Hole in a plane : (a) A hole of radius \(R\) is cut out from a very large flat sheet with uniform charge density \(\sigma\). Let \(L\) be the line perpendicular to the sheet, passing through the center of the hole. What is the electric field at a point on \(L\), a distance \(z\) from the center of the hole? Hint: Consider the plane to consist of many concentric rings. (b) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), very close to the center of the hole, show that it undergoes oscillatory motion, and find the frequency \(\omega\) of these oscillations. What is \(\omega\) if \(m=1 \mathrm{~g},-q=-10^{-8} \mathrm{C}, \sigma=10^{-6} \mathrm{C} / \mathrm{m}^{2}\), and \(R=0.1 \mathrm{~m} ?\) (c) If a charge \(-q\) with mass \(m\) is released from rest on \(L\), a distance \(z\) from the sheet, what is its speed when it passes through the center of the hole? What does your answer reduce to for large \(z\) (or, equivalently, small \(R\) )?

Zero force from a triangle ** Two positive ions and one negative ion are fixed at the vertices of an equilateral triangle. Where can a fourth ion be placed, along the symmetry axis of the setup, so that the force on it will be zero? Is there more than one such place? You will need to solve something numerically.

Field from a semicircle * A thin plastic rod bent into a semicircle of radius \(R\) has a charge \(Q\) distributed uniformly over its length. Find the electric field at the center of the semicircle.

Oscillating in a ring A ring with radius \(R\) has uniform positive charge density \(\lambda\). A particle with positive charge \(q\) and mass \(m\) is initially located at the center of the ring and is then given a tiny kick. If it is constrained to move in the plane of the ring, show that it undergoes simple harmonic motion (for small oscillations), and find the frequency. Hint: Find the potential energy of the particle when it is at a (small) radius, \(r\), by integrating over the ring, and then take the negative derivative to find the force. You will need to use the law of cosines and also the Taylor series \(1 / \sqrt{1+\epsilon} \approx 1-\epsilon / 2+3 \epsilon^{2} / 8\)

Potential energy of a cylinder * * Problem \(1.24\) gives one way of calculating the energy per unit length stored in a solid cylinder with radius \(a\) and uniform volume charge density \(\rho .\) Calculate the energy here by using Eq. (1.53) to find the total energy per unit length stored in the electric field. Don't forget to include the field inside the cylinder. You will find that the energy is infinite, so instead calculate the energy relative to the configuration where all the charge is initially distributed uniformly over a hollow cylinder with large radius \(R\). (The field outside radius \(R\) is the same in both configurations, so it can be ignored when calculating the relative energy.) In terms of the total charge \(\lambda\) per unit length in the final cylinder, show that the energy per unit length can be written as \(\left(\lambda^{2} / 4 \pi \epsilon_{0}\right)(1 / 4+\ln (R / a))\).
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