Question

\(A\) total charge of \(Q\) is distributed uniformly on a line segment of length \(2 L\) along the \(y\) -axis (see figure). The \(x\) -component of the electric field at a point \((a, 0)\) on the \(x\) -axis is given by $$E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}}$$ where \(k\) is a physical constant and \(a>0\) a. Confirm that \(E_{x}(a)=\frac{k Q}{a \sqrt{a^{2}+L^{2}}}\) b. Letting \(\rho=Q / 2 L\) be the charge density on the line segment, show that if \(L \rightarrow \infty,\) then \(E_{x}(a)=2 k \rho / a\)

Short answer

In part a, we confirmed that the given integral expression for the x-component of the electric field, \(E_x(a)\), simplifies to the formula \(E_x(a) = \frac{kQ}{a\sqrt{a^2+L^2}}\). In part b, by taking the limit as L approaches infinity, we showed that \(E_x(a)\) becomes \(\frac{2k\rho}{a}\), where \(\rho = \frac{Q}{2L}\) is the charge density.

Step-by-step solution

Part a: Solving the integral

We are given the expression for the x-component of the electric field, \(E_x(a)\), as: $$E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}}$$ To solve this integral, we substitute \(y = a\tan(\theta)\), and \(dy = a\sec^2(\theta)d\theta\): $$E_{x}(a) = \frac{k Qa}{2L} \int_{-\arctan(\frac{L}{a})}^{\arctan(\frac{L}{a})} \frac{a\sec^2(\theta)}{\left(a^2 + a^2\tan^2(\theta)\right)^{3/2}} d\theta$$ Simplify the expression by cancelling out \(a\) and combining terms: $$E_{x}(a) = \frac{k Q}{2 L} \int_{-\arctan(\frac{L}{a})}^{\arctan(\frac{L}{a})} \frac{\sec^2(\theta)}{(a^2\sec^2(\theta))^{3/2}} d\theta$$ Using the identity \(\sec^2(\theta) = 1 + \tan^2(\theta)\), we have: $$E_{x}(a) = \frac{k Q}{2 L} \int_{-\arctan(\frac{L}{a})}^{\arctan(\frac{L}{a})} \frac{\sec^2(\theta)}{(a^2(1+\tan^2(\theta)))^{3/2}} d\theta$$ Now, the expression simplifies to: $$E_{x}(a) = \frac{k Q}{2 L} \int_{-\arctan(\frac{L}{a})}^{\arctan(\frac{L}{a})} d\theta = \frac{kQ}{2L} \left[\arctan(\frac{L}{a}) - (-\arctan(\frac{L}{a}))\right]$$ This further simplifies to: $$E_{x}(a) = \frac{kQ}{a\sqrt{a^2+L^2}}$$ which confirms the given expression.

Part b: Taking the limit as length tends to infinity

We are given the charge density as, \(\rho = \frac{Q}{2L}\) and need to show that when \(L \rightarrow \infty\), \(E_{x}(a) = \frac{2k\rho}{a}\). First, we express \(E_{x}(a)\) in terms of charge density: $$E_{x}(a) = \frac{kQ}{a\sqrt{a^2+L^2}} = k(2L\rho)\frac{1}{a\sqrt{a^2+L^2}}$$ Now we take the limit as \(L \rightarrow \infty\): $$\lim_{L \to \infty} E_{x}(a) = \lim_{L \to \infty} \frac{2kL\rho}{a\sqrt{a^2+L^2}}$$ To evaluate this limit, we can divide the numerator and denominator by \(L\): $$\lim_{L \to \infty} \frac{2kL\rho}{a\sqrt{a^2+L^2}} = \lim_{L \to \infty} \frac{2k\rho}{\frac{a}{L}\sqrt{1+\frac{a^2}{L^2}}}$$ In the limit as \(L\) approaches infinity, \(\frac{a}{L}\) tends to zero, thus we can simplify the expression: $$\lim_{L \rightarrow \infty} \frac{2k\rho}{\frac{a}{L}\sqrt{1+\frac{a^2}{L^2}}} = \frac{2k\rho}{0\sqrt{1+0}}$$ Now, we have the desired expression: $$E_{x}(a) = \frac{2k\rho}{a}$$ when \(L \rightarrow \infty\).

Key concepts

Charge Distribution

Charge distribution refers to the way electric charge is spread out in a material or space. In the context of an electric field, understanding how the charge is distributed helps in calculating the field's influence at different points.
In the exercise, the total charge, denoted as \(Q\), is distributed uniformly along a line segment of length \(2L\) on the \(y\)-axis. This means the charge density, or the amount of charge per unit length, is constant across the segment. Uniform charge distribution is significant because it simplifies the calculation of electric fields, as the charge density \( \rho \) can be easily defined as \( \rho = \frac{Q}{2L} \).
When charges are uniformly distributed, the effects of the electric field can be predicted more accurately, particularly when dealing with symmetrical geometries. Such predictability is key in solving integrals and understanding the limits, which are central to the given exercise.

Line Integral

A line integral in this context is a way to sum up the electric field components along a path. Essentially, it measures how a vector field, like the electric field, accumulates over a certain curve or path.
The exercise makes use of a line integral to compute the \(x\)-component of the electric field due to a charge distribution on the \(y\)-axis. The integral \(E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}} \)is evaluated over the length of the charge distribution. The integral expression represents how each infinitesimal piece of charge on the \(y\)-axis contributes to the electric field at a particular point on the \(x\)-axis.
This type of integral requires substitutions and simplifications, like those using trigonometric identities, to evaluate correctly. Such techniques are critical as they transform complex expressions into more manageable forms.

Limits in Calculus

In calculus, limits help understand the behavior of functions as inputs approach a certain value. They are essential for evaluating expressions that lead towards infinity or zero, as they often do in physical contexts.
In part b of the solution, limits are used to determine the behavior of the electric field \(E_{x}(a)\) as the length \(L\) of the charged segment tends to infinity. The expression \(\lim_{L \to \infty} E_{x}(a)\) simplifies due to dividing both numerator and denominator by \(L\), leading to insights into how the electric field behaves under an infinitely large charge distribution.
This technique is indicative of how limits allow physicists and engineers to assess conditions that are challenging to measure directly, providing a theoretical framework to predict behavior in idealized circumstances.

Uniform Charge Density

Uniform charge density describes a situation where charge per unit length, area, or volume is constant throughout the distribution. This simplifies calculations as it implies a symmetric field impact.
In the exercise, the charge density is given by \( \rho = \frac{Q}{2L} \), implying a linear charge distribution along the \(y\)-axis. Uniform charge density is crucial for applying Coulomb's law and performing line integrals, reducing complex real-world problems into manageable mathematical expressions.

• This assumes charge distribution is evenly spread, which is often a valid approximation in many physics problems.
• It highlights symmetry—vital for evaluating integrals and electric fields, especially when symmetry can be exploited to simplify integrals or set boundary conditions.

This concept is foundational for deriving expressions that describe electric fields due to extended objects, making uniform distributions a common starting point in electrostatics.