/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 A thin rod lies on the \(x\) -ax... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 57

A thin rod lies on the \(x\) -axis between \(x=0\) and \(x=L\) and carries total charge \(Q\) distributed uniformly over its length. Show that the electric field strength for \(x>L\) is given by \(E=k Q /[x(x-L)]\)

Short Answer

Expert verified
The electric field strength for x>L is indeed given by \(E=\frac{k~Q}{x(x-L)}\).

Step by step solution

01

Define the Electric Field Contribution

The infinitesimal amount of charge \(dq\) located at position \(x'\) from the origin creates an electric field \(d\vec{E}\) at a position \(x\) (with \(x>x'\)) defined by \(d\vec{E}=\frac{k~dq}{(x-x')^2}\hat{i}\) directed along the positive \(x\)-axis. The infinitesimal charge \(dq\) is given by \(dq=位dx'=Q/L~dx'\), where \(位=Q/L\) is the linear charge density over the rod.
02

Integrate Over the Length of the Rod

The total electric field at position \(x\), resulting from the superposition of the electric field contributions from all charge elements, is given by the integral over all \(d\vec{E}\): \(E=\int_{0}^{L}d\vec{E}\). Substituting the expressions for \(d\vec{E}\) and \(dq\), we obtain: \(E=k~位\int_{0}^{L}\frac{dx'}{(x-x')^2}\).\nMake sure to integrate correctly, pay attention to the limits of the integral (which correspond to the positions of the endpoints of the rod on the \(x'\)-axis) and handle the integral of the reciprocal square function correctly.
03

Evaluate the Integral

Evaluating the integral yields: \(E=k~位[-\frac{1}{x-x'}]_{0}^{L}=k~位[\frac{1}{x-L}-\frac{1}{x}]\). Substituting back for \(位=Q/L\), we obtain: \(E=k~Q[\frac{1}{L}(\frac{1}{x-L}-\frac{1}{x})]=k~Q[\frac{x-(x-L)}{x(x-L)}]\), which simplifies to \(E=\frac{k~Q}{x(x-L)}\) as required to show.

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