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Chapter 21: Problem 10

The field of an infinite charged line decreases as \(1 / r .\) Why isn't this a violation of the inverse-square law?

Short Answer

Expert verified
It's not a violation of the inverse-square law because that law applies to point sources in three dimensions. An infinitely charged line, being essentially one-dimensional, will have its field decrease linearly with distance, which is \(1/r\), and not as \(1/r^2\). This is due to the fact that the charge spreads out over the 2D surface of a cylinder that encapsulates the line.

Step by step solution

01

Understanding the Inverse-Square Law

The inverse-square law suggests that for a point source in three dimensions, the intensity (or influence) of the source decreases as the square of the distance from the source. Mathematically, it's expressed as \(I \propto 1/r^2\). However, this law applies to point sources radiating out in three dimensions.
02

Consider Dimensionality

The exercise gives an infinite line of charge. This line is extending in one dimensional space. Therefore, the charge from the line does not dissipate in all directions equally. Instead, it spreads out in a cylindrical shell around the line charge. We need to take into account the linear charge distribution of this one-dimensional entity.
03

Applying the Principle

The electric field of an infinite line charge decreases linearly with the distance i.e., it goes as \(1/r\). This is because the charge from the line spreads out over the surface of a cylinder with radius \(r\), leading to a field strength proportional to \(1/r\) along the surface of the cylinder. This perfectly aligns with the two-dimensional character of a cylinder surface.

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Most popular questions from this chapter

What's the approximate field strength \(1 \mathrm{cm}\) above a sheet of paper carrying uniform surface charge density \(\sigma=45 \mathrm{nC} / \mathrm{m}^{2} ?\)

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