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Chapter 1: Problem 23

Field near a stick A stick with length \(2 \ell\) has uniform linear charge density \(\lambda\). Consider a point \(P\), a distance \(\eta \ell\) from the center (where \(0 \leq \eta<1\) ), and an infinitesimal distance away from the stick. Up close, the stick looks infinitely long, as far as the \(\mathrm{E}\) component perpendicular to the stick is concerned. So we have \(E_{\perp}=\lambda / 2 \pi \epsilon_{0} r\). Find the E component parallel to the stick, \(E_{I}\). Does it approach infinity, or does it remain finite at the end of the stick?

Short Answer

Expert verified
The electric field component \(E_{I}\) parallel to the charged stick approaches infinity at the end of the stick when calculated for a uniformly charged stick of length \(2 \ell\) and uniform linear charge density \( \lambda \).

Step by step solution

01

Start with the expression for the electric field

The expression for the electric field of an infinitely long line of charge is given by \(E_{\perp}=\lambda / 2 \pi \epsilon_{0} r\), where \( \lambda \) is the linear charge density and r is the perpendicular distance from the line charge.
02

Calculate the x-component of the electric field

For an infinitesimal charge \( dq = \lambda dx \), located at a distance x from point \(P\), we can describe the electric field \( d\vec{E} \) due to \( dq \) at \( P \). The x-component is \( dE_x = \frac{k dq cos(\theta)}{r^2} = \frac{k \lambda dx cos(\theta)}{r^2} \). \( cos(\theta) \) is the angle between the x direction and \( \vec{r} \), which can be found using the ratios of sides in right triangles. It can be calculated to be \( \frac{x}{r} \) where \( r = \sqrt{x^2 + \eta^2 l^2} \). Therefore, we have \( dE_x = k \lambda dx \frac{x}{x^2 + \eta^2 l^2} \).
03

Integrate over the length of the stick

We need to evaluate the total E-field along the axis of the uniformly charged stick. We do this by integrating from \( -l \) to \( l \) because point P is at a distance \( \eta l \) from the center of the stick. So, the total x-component of the electric field is \( E_x = k \lambda \int_{-l}^{+l} dx \frac{x}{x^2 + \eta^2 l^2} \).
04

Evaluate the integral

The integral can be expressed in terms of arctan function. Four key steps involved are performing a substitution, differentiating, inserting limits, and performing the final calculation. Finally, after simplifying, the integration leads us to \( E_x = \frac{\lambda}{2 \pi \epsilon_{0} \eta l} [ arctan(\frac{1+\eta}{\eta}) - arctan(\frac{1-\eta}{\eta})] \)
05

Check the behaviour of \(E_x\) as \( \eta \rightarrow 1\)

As \( \eta \rightarrow 1 \), \( E_x \rightarrow \frac{\lambda}{2 \pi \epsilon_{0} l} \times \infty \), therefore, the E component parallel to the stick, \(E_{I}\) approaches infinity at the end of the stick.

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Key Concepts

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

Linear Charge Density
Linear charge density, denoted by \( \lambda \), is a measure of the amount of electric charge per unit length along a line. Think of it as if you were spreading a fixed amount of charge evenly over a piece of string; the denser the charge along the string, the greater the linear charge density would be. This concept is crucial when dealing with the electric fields created by charged objects that are very long compared to their other dimensions, such as wires, sticks, or rods.

In the context of our exercise, the stick with length \(2 \ell\) has uniform linear charge density \(\lambda\), which implies that the stick carries the same distribution of charge all throughout its length. The symmetry due to this uniform distribution allows us to use certain assumptions and simplifications when calculating the electric field, particularly as we move to address the electric field components around the stick.
Electric Field Components
The electric field, represented by \(\vec{E}\), is a vector quantity, meaning it has both magnitude and direction. It can be decomposed into components to understand the effect of an electric charge in different directions. In our problem involving a charged stick, we are particularly interested in two components: \(E_\perp\) and \(E_I\), which are the electric field components perpendicular and parallel to the length of the stick, respectively.

The \(E_\perp\) component is straightforward to calculate for an infinitely long charged line, as it only depends on the linear charge density and the distance from the charge. However, for the parallel component, \(E_I\), we must look at the contributions of all the infinitesimal charged elements \(dq\) along the stick. This is typically done via integration to sum up the influence of each small charge element over the length of the stick. The behavior of these components provides insights into how the electric field varies around the charged object.
Integration of Electric Field
Integration is a powerful mathematical tool used to sum up infinitely many infinitesimal contributions to a whole. In the context of electric fields, integration allows us to calculate the total electric field produced by a continuous distribution of charge, rather than just considering a single point charge.

Returning to our example, the calculation of the electric field component parallel to the stick, \(E_I\), requires us to integrate the influence of each differential element of charge \(dq\) along the stick. By setting up the integral over the length of the stick from \( -\ell \) to \( \ell \) and integrating, we incorporate the effects of all charge elements. This process highlights the importance of integration in understanding the behavior of electric fields in more complex charge distributions. The integral results in an expression involving the arctan function, which describes \(E_I\) in terms of the stick's charge density and its distance from the point of interest.

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Most popular questions from this chapter

Ball in a sphere We know that if a point charge \(q\) is located at radius \(a\) in the interior of a sphere with radius \(R\) and uniform volume charge density \(\rho\), then the force on the point charge is effectively due only to the charge that is located inside radius \(a\). (a) Consider instead a uniform ball of charge located entirely inside a larger sphere of radius \(R\). Let the ball's radius be \(b\), and let its center be located at radius \(a\) in the larger sphere. Its volume charge density is such that its total charge is \(q\). Assume that the ball is superposed on top of the sphere, so that all of the sphere's charge is still present. Can the force on the ball be obtained by treating it like a point charge and considering only the charge in the larger sphere that is inside radius \(a\) ? (b) Would the force change if we instead remove the charge in the larger sphere where the ball is? So now we are looking at the force on the ball due to the sphere with a cavity carved out, which is a more realistic scenario.

An equilateral triangle Three positive charges, \(A, B\), and \(C\), of \(3 \cdot 10^{-6}, 2 \cdot 10^{-6}\), and \(2 \cdot 10^{-6}\) coulombs, respectively, are located at the corners of an equilateral triangle of side \(0.2 \mathrm{~m}\). (a) Find the magnitude in newtons of the force on each charge. (b) Find the magnitude in newtons/coulomb of the electric field at the center of the triangle.

Hydrogen atom \(* *\) The neutral hydrogen atom in its normal state behaves, in some respects, like an electric charge distribution that consists of a point charge of magnitude \(e\) surrounded by a distribution of negative charge whose density is given by \(\rho(r)=-C e^{-2 r / a_{0}} .\) Here \(a_{0}\) is the Bohr radius, \(0.53 \cdot 10^{-10} \mathrm{~m}\), and \(C\) is a constant with the value required to make the total amount of negative charge exactly \(e\). What is the net electric charge inside a sphere of radius \(a_{0} ?\) What is the electric field strength at this distance from the nucleus?

Zero field inside a cylindrical shell * Consider a distribution of charge in the form of a hollow circular cylinder, like a long charged pipe. In the spirit of Problem 1.17, show that the electric field inside the pipe is zero.

Deriving the energy density ** Consider the electric field of two protons a distance \(b\) apart. According to Eq. (1.53) (which we stated but did not prove), the potential energy of the system ought to be given by $$ \begin{aligned} U &=\frac{\epsilon_{0}}{2} \int \mathbf{E}^{2} d v=\frac{\epsilon_{0}}{2} \int\left(\mathbf{E}_{1}+\mathbf{E}_{2}\right)^{2} d v \\ &=\frac{\epsilon_{0}}{2} \int \mathbf{E}_{1}^{2} d v+\frac{\epsilon_{0}}{2} \int \mathbf{E}_{2}^{2} d v+\epsilon_{0} \int \mathbf{E}_{1} \cdot \mathbf{E}_{2} d v \end{aligned} $$ where \(\mathbf{E}_{1}\) is the field of one particle alone and \(\mathbf{E}_{2}\) that of the other. The first of the three integrals on the right might be called the "electrical self-energy" of one proton; an intrinsic property of the particle, it depends on the proton's size and structure. We have always disregarded it in reckoning the potential energy of a system of charges, on the assumption that it remains constant; the same goes for the second integral. The third integral involves the distance between the charges. Evaluate this integral. This is most easily done if you set it up in spherical polar coordinates with one of the protons at the origin and the other on the polar axis, and perform the integration over \(r\) before the integration over \(\theta\). Thus, by direct calculation, you can show that the third integral has the value \(e^{2} / 4 \pi \epsilon_{0} b\), which we already know to be the work required to bring the two protons in from an infinite distance to positions a distance \(b\) apart. So you will have proved the correctness of Eq. (1.53) for this case, and by invoking superposition you can argue that Eq. (1.53) must then give the energy required to assemble any system of charges.
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