/*! 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 125 The energy density in the electr... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 125

The energy density in the electric field created by a point charge falls off with distance from the point charge as (A) \(\frac{1}{r}\) (B) \(\frac{1}{r^{2}}\) (C) \(\frac{1}{r^{3}}\) (D) \(\frac{1}{r^{4}}\)

Short Answer

Expert verified
The energy density in the electric field created by a point charge falls off with distance from the point charge as \(\frac{1}{r^4}\) (option D). This can be determined by substituting the formula for the electric field due to a point charge, \(\frac{kQ}{r^2}\), into the formula for energy density, \(\frac{1}{2} \epsilon_0 E^2\), and simplifying the expression.

Step by step solution

01

Recall the formula for electric field due to a point charge

The electric field, \(E\), due to a point charge, \(Q\), at a distance \(r\) from the charge is given by Coulomb's Law: \[E = \frac{kQ}{r^2}\] where \(k\) is the electrostatic constant, approximately \(8.99 \times 10^{9} \frac{N m^2}{C^2}\).
02

Recall the formula for energy density in an electric field

The energy density, \(u\), in an electric field, is given by the following formula: \[u = \frac{1}{2} \epsilon_0 E^2\] where \(\epsilon_0\) is the vacuum permittivity, approximately \(8.85 \times 10^{-12} \frac{C^2}{N m^{2}}\), and \(E\) is the electric field.
03

Substitute the formula for the electric field in the energy density formula

We substitute the expression for the electric field due to a point charge into the formula for energy density: \[u = \frac{1}{2} \epsilon_0 \left(\frac{kQ}{r^2}\right)^2\]
04

Simplify the expression for energy density

Now, simplify the expression by squaring the electric field term and combining constants: \[u = \frac{1}{2} \epsilon_0 \frac{k^2 Q^2}{r^4}\]
05

Determine how the energy density falls off with distance

We can see from the simplified expression that the energy density, \(u\), falls off with distance, \(r\), as \(\frac{1}{r^4}\). Therefore, the correct answer is (D) \(\frac{1}{r^{4}}\).

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