Assume that a stationary electron is a point of charge.

(a)What is the energy density of its electric field at radial distances r=1.00mm ?

(b)What is the energy density of its electric field at radial distancesr=1.00μm?

(c)What is the energy density of its electric field at radial distancesr=1.00nm?

(d)What is the energy density of its electric field at radial distances r=1.00pm?

(e)What is uin the limit as r0?

Short Answer

Expert verified
  1. Energy density at radial distance r=1.00mmis 9.16x10-18J/m3.
  2. Energy density at radial distancer=1.00μmis9.16x10-6J/m3.
  3. Energy density at radial distancer=1.00nmis9.16x106J/m3.
  4. Energy density at radial distance r=1.00pmis 9.16x1018J/m3.
  5. Energy density in the limit as r0is .

Step by step solution

01

The given data

  1. A stationary point charge of electron is assumed.
  2. Distance,r=1.00mm
  3. Distance,r=1.00μm
  4. Distance,r=1.00nm
  5. Distance,r=1.00pm
02

Understanding the concept of the energy density

The quantity of energy that is contained in a system or area of space per unit volume is known as the energy density. We use the equation of energy density stored in the electrostatic field. Substituting the value of the magnitude of the electric field in density, we can calculate the energy density for the different radial distances.

Formulae:

The energy stored per unit volume that is energy density ( u ) between the capacitors,

u=12ε0E2 …(i)

The electric field at a point due to stored charge, E=q4πε0r2 …(ii)

Where,

q is charge of electron q=1.6×10-19C

ε0=8.85×10-12C2N.m2 is permittivity of free space

03

(a) Calculation of the energy density at  r = 1 mm

For the formula of energy density due to electric field, we substitute the value of equation (ii) in equation (i) as follows:

u=12ε0q4πε0r22=q32π2ε0r4

…(iii)

Substituting all the given values in equation (iii), we get the energy density at r = 1.0 mm as follows:

u=1.6×10-19C232π2×8.85×10-12C2N.m2×10-3m4=2.56×10-38C22.795×10-21C2.m2N=9.16×10-18J/m3

Hence, the value of the energy stored per volume is 9.16×10-18J/m3.

04

(b) Calculation of the energy density at r=1 μm

Substituting all the given values in equation (a), we get the energy density at r=1.00μmas follows:

u=1.6×10-19C232π2×8.85×10-12C2N.m2×10-64=2.56×10-38C22.795×10-33C2.m2N=9.16×10-6J/m3

Hence, the value of the energy stored per volume is 9.16×10-6J/m3.

05

(c) Calculation of the energy density at r = 1 nm

Substituting all the given values in equation (a), we get the energy density at r = 1.00 nm as follows:

u=1.6×10-19C232π2×8.85×10-12C2N.m2×10-94=2.56×10-38C22.795×10-45C2.m2N=9.16×106J/m3

Hence, the value of the energy stored per volume is 9.16×10-6J/m3.

06

(d) Calculation of the energy density at r = 1 pm

Substituting all the given values in equation (a), we get the energy density at r = 1 pm as follows:

u=1.6×10-19C232π2×8.85×10-12C2N.m2×10-124=2.56×10-382.795×10-57=9.16×1018J/m3

Hence, the value of the energy stored per volume is 9.16×1018J/m3.

07

(e) Calculation of the energy density

From equation (a), we can get the relation of energy density to distance as follows:

ur-4

As the value of r approaches r0, the value of the energy density is .

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free