Question

Assume that a stationary electron is a point of charge.What is the energy density u of its electric field at radial distances (a) r ! 1.00 mm, (b) r ! 1.00 mm, (c) r ! 1.00 nm, and (d) r ! 1.00 pm

Answer 1

See explanation.

Step-by-step explanation:

The energy density
`U_E` of an electric field
`E` is given by

`U_E = (1)/(2) \varepsilon_0E^2.`

The electric field
`E` at a distance
`r` due to the electron is

`E = (q_e)/(4\pi \varepsilon_0r^2 );`

therefore,

`U_E = (q_e^2)/(32\pi^2 \varepsilon_0r^4 ).`

putting in values for
`q_e = 1.6*10^(-19)C` and
`\varepsilon_0 = 8.85*10^(-12)C^2/N\cdot m^2` we get:

`U_E = (9.16*10^(-30))/(r^4)`

(a).

For
`r=1.00mm`

`U_E = (9.16*10^(-30))/((1*10^(-3))^4)`

`\boxed{U_E = 9.16^(-18)J/m^3}`

(b)

For
`r=1.00mm`

`U_E = (9.16*10^(-30))/((1*10^(-3))^4)`

`\boxed{U_E = 9.16^(-18)J/m^3}`

(c).

For
`r =1.00nm`

`U_E = (9.16*10^(-30))/((1*10^(-9))^4)`

`\boxed{U_E = 9.16*10^6J/m^3}`

(d).

For
`r = 1.00pm`

`U_E = (9.16*10^(-30))/((1*10^(-12))^4)`

`\boxed{ U_E = 9.16*10^(18)J/m^3}`