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Chapter 4: Problem 18

Find the potential at the origin produced by a line charge \(\rho_{L}=k x /\left(x^{2}+a^{2}\right)\) extending along the \(x\) axis from \(x=a\) to \(+\infty\), where \(a>0\). Assume a zero reference at infinity.

Short Answer

Expert verified
Answer: The potential at the origin is \(V(\mathbf{0}) = \frac{k a \pi}{16 \epsilon_{0}}\).

Step by step solution

01

Write down the formula for electric potential due to a continuous charge distribution

We will use the formula for electric potential due to a continuous charge distribution: \(V(\mathbf{r}) = \int \frac{1}{4\pi \epsilon_{0}} \frac{\rho_{L}(\mathbf{r}')}{|\mathbf{r} - \mathbf{r'}|} d\mathbf{r'}\) Here, \(\epsilon_{0}\) is the vacuum permittivity, \(\rho_{L}\) is the linear charge density, \(\mathbf{r}\) is the position where we want to calculate the potential, and \(\mathbf{r'}\) represents positions on the line charge.
02

Set up the integral

We have \(\rho_{L} = k x /\left(x^{2}+a^{2}\right)\) and we want to find the potential at the origin, so \(\mathbf{r} = \mathbf{0}\). Since the line charge extends along the \(x\) axis, we have \(\mathbf{r'} = x'\mathbf{\hat{x}}\). Therefore, \(|\mathbf{r} - \mathbf{r'}| = |-\mathbf{r'}| = |x'|\mathbf{\hat{x}}\). We can now set up the integral: \(V(\mathbf{0}) = \int_{a}^{\infty} \frac{1}{4\pi \epsilon_{0}} \frac{k x'}{x'^{2}+a^{2}} \frac{1}{|x'|\mathbf{\hat{x}}} dx'\)
03

Simplify the integral

Now, we can simplify the integral by canceling the \(x'\) terms: \(V(\mathbf{0}) = \frac{k}{4\pi \epsilon_{0}} \int_{a}^{\infty} \frac{1}{x'^{2}+a^{2}} dx'\)
04

Evaluate the integral

This is a standard integral of the form \(\int_{a}^{\infty} \frac{1}{x'^{2}+a^{2}} dx'\). To evaluate it, we can use the substitution \(u = x'/a\), so \(x' = au\) and \(dx' = a du\). The new limits of integration are \(u = 1\) and \(u = \infty\). The integral becomes: \(V(\mathbf{0}) = \frac{k}{4\pi \epsilon_{0}} \int_{1}^{\infty} \frac{a}{a^{2}u^{2} + a^{2}} a du \) \(V(\mathbf{0}) = \frac{k}{4\pi \epsilon_{0}} a \int_{1}^{\infty} \frac{1}{u^{2} + 1} du\) Now, this is a well-known integral, and its solution is \(\arctan{u}\). Therefore, we can evaluate the result as: \(V(\mathbf{0}) = \frac{k a}{4\pi \epsilon_{0}} \left[\arctan{u}\right]_{1}^{\infty}\)
05

Calculate the potential at the origin

Applying the limits of integration, we get: \(V(\mathbf{0}) = \frac{k a}{4\pi \epsilon_{0}} \left[\lim_{u \to \infty} (\arctan{u}) - \arctan{1}\right]\) Since \(\lim_{u \to \infty} (\arctan{u}) = \frac{\pi}{2}\) and \(\arctan{1} = \frac{\pi}{4}\), we find: \(V(\mathbf{0}) = \frac{k a}{4\pi \epsilon_{0}} \left[\frac{\pi}{2} - \frac{\pi}{4}\right]\) \(V(\mathbf{0}) = \frac{k a}{4\pi \epsilon_{0}} \frac{\pi}{4}\) Finally, we have the potential at the origin: \(V(\mathbf{0}) = \frac{k a \pi}{16 \epsilon_{0}}\)

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Most popular questions from this chapter

A positive point charge of magnitude \(q_{1}\) lies at the origin. Derive an expression for the incremental work done in moving a second point charge \(q_{2}\) through a distance \(d x\) from the starting position \((x, y, z)\), in the direction of \(-\mathbf{a}_{x}\)

Two point charges, \(1 \mathrm{nC}\) at \((0,0,0.1)\) and \(-1 \mathrm{nC}\) at \((0,0,-0.1)\), are in free space. \((a)\) Calculate \(V\) at \(P(0.3,0,0.4) .\) (b) Calculate \(|\mathbf{E}|\) at \(P .(c)\) Now treat the two charges as a dipole at the origin and find \(V\) at \(P\).

A line charge of infinite length lies along the \(z\) axis and carries a uniform linear charge density of \(\rho_{\ell} \mathrm{C} / \mathrm{m}\). A perfectly conducting cylindrical shell, whose axis is the \(z\) axis, surrounds the line charge. The cylinder (of radius \(b\) ), is at ground potential. Under these conditions, the potential function inside the cylinder \((\rhob .(d)\) Find the stored energy in the electric field per unit length in the \(z\) direction within the volume defined by \(\rho>a\), where \(a

In a certain medium, the electric potential is given by $$ V(x)=\frac{\rho_{0}}{a \epsilon_{0}}\left(1-e^{-a x}\right) $$ where \(\rho_{0}\) and \(a\) are constants. ( \(a\) ) Find the electric field intensity, E. \((b)\) Find the potential difference between the points \(x=d\) and \(x=0 .(c)\) If the medium permittivity is given by \(\epsilon(x)=\epsilon_{0} e^{a x}\), find the electric flux density, \(\mathbf{D}\), and the volume charge density, \(\rho_{v}\), in the region. ( \(d\) ) Find the stored energy in the region \((0

Uniform surface charge densities of 6 and \(2 \mathrm{nC} / \mathrm{m}^{2}\) are present at \(\rho=2\) and \(6 \mathrm{~cm}\), respectively, in free space. Assume \(V=0\) at \(\rho=4 \mathrm{~cm}\), and calculate \(V\) at \((a) \rho=5 \mathrm{~cm} ;(b) \rho=7 \mathrm{~cm} .\)
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