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Chapter 23: Problem 34

Consider a proton to be a uniformly charged sphere 1 fm in radius. Find the electric energy density at the proton's surface.

Short Answer

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To get the final solution, calculate all the above steps properly.

Step by step solution

01

Understand the Basics

Recognize a proton as a uniformly charged sphere with radius 1 femtometer (fm). The charge of a proton is \(+e\) where \(e = 1.6 \times 10^{-19} C\). The electric field \(E\) just outside the surface of the charged sphere can be given by Coulomb's Law: \(E = k \frac{Q}{r^2}\), where \(Q\) is the charge of the sphere, \(r\) is the radius of the sphere and \(k\) is Coulomb's constant.
02

Calculate the Electric Field

Plug the values of \(Q\), \(r\) and \(k\) into the formula. In this case, \(Q = e = 1.6 \times 10^{-19} C\), \(r = 1 \times 10^{-15} m\) (converted from fm) and \(k = 9 \times 10^9 N m^2/C^2\). Calculate the electric field \(E\) using these values.
03

Calculate the Energy Density

The electrostatic energy density \(u\) in a region of space with electric field \(E\) is given by \(u = \frac{1}{2} \epsilon_0 E^2\), where \(\epsilon_0\) is the permittivity of free space. Substituting the calculated \(E\) value and \(\epsilon_0 = 8.85 \times 10^{-12} C^2/N m^2\) into the equation, calculate \(u\), the electric energy density at the proton's surface.

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Most popular questions from this chapter

Does the superposition principle hold for electric-field energy densities? That is, if you double the field strength at some point. do you double the energy density as well?

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