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Chapter 23: Problem 68

A classical view of the electron pictures it as a purely electric entity, whose Einstein rest mass energy, \(E=m c^{2},\) is the energy stored in its electric field. If the electron were a sphere with charge distributed uniformly over its surface, what radius would it have in order to satisfy this condition? (Note: Your answer, and the picture of the electron as a sphere, aren't consistent with quantum theory.)

Short Answer

Expert verified
After substituting the known values, the radius, \(r\), of an electron treated as a sphere with uniformly distributed charge over its surface becomes approximately \(2.818 \times 10^{-15}\) m.

Step by step solution

01

Write Down Known Variables

The charge of an electron is \(1.602 \times 10^{-19}\) C and its mass is \(9.11 \times 10^{-31}\) kg. The speed of light, \(c\), is \(2.998 \times 10^{8}\) m/s and the permittivity of free space, \(\varepsilon_{0}\), is \(8.85 \times 10^{-12}\) C\(^2\)/Nm\(^2\).
02

Write Down The Equation of Energy Stored in an Electric Field

We can find the energy, \(E\), stored in an electric field around a spherical charge, \(Q\), by the formula, \(E = \frac{Q^2}{8 \pi \varepsilon_{0} r}\). Here, \(r\) represents the radius of the sphere.
03

Write Down The Equation for Einstein's Rest Mass Energy

We have the rest mass energy equation, \(E = m c^{2}\).
04

Equate The Two Forms of Energy

We assume that the electron's rest mass energy is completely stored in its electric field. Therefore, we can equate the two expressions for energy: \(\frac{Q^2}{8 \pi \varepsilon_{0} r} = m c^{2}\).
05

Solve for Radius

We need to solve this equation for the radius, \(r\). We get \(r = \frac{Q^2}{8 \pi \varepsilon_{0} m c^{2}}\). Substituting the known values, we find the radius of the sphere.

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Most popular questions from this chapter

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