Answer:
See explanation.
Explanation:
The energy density [tex]U_E[/tex] of an electric field [tex]E[/tex] is given by
[tex]U_E = \dfrac{1}{2} \varepsilon_0E^2.[/tex]
The electric field [tex]E[/tex] at a distance [tex]r[/tex] due to the electron is
[tex]E = \dfrac{q_e}{4\pi \varepsilon_0r^2 };[/tex]
therefore,
[tex]U_E = \dfrac{q_e^2}{32\pi^2 \varepsilon_0r^4 }.[/tex]
putting in values for [tex]q_e = 1.6*10^{-19}C[/tex] and [tex]\varepsilon_0 = 8.85*10^{-12}C^2/N\cdot m^2[/tex] we get:
[tex]U_E = \dfrac{9.16*10^{-30}}{r^4}[/tex]
(a).
For [tex]r=1.00mm[/tex]
[tex]U_E = \dfrac{9.16*10^{-30}}{(1*10^{-3})^4}[/tex]
[tex]\boxed{U_E = 9.16^{-18}J/m^3}[/tex]
(b)
For [tex]r=1.00mm[/tex]
[tex]U_E = \dfrac{9.16*10^{-30}}{(1*10^{-3})^4}[/tex]
[tex]\boxed{U_E = 9.16^{-18}J/m^3}[/tex]
(c).
For [tex]r =1.00nm[/tex]
[tex]U_E = \dfrac{9.16*10^{-30}}{(1*10^{-9})^4}[/tex]
[tex]\boxed{U_E = 9.16*10^6J/m^3}[/tex]
(d).
For [tex]r = 1.00pm[/tex]
[tex]U_E = \dfrac{9.16*10^{-30}}{(1*10^{-12})^4}[/tex]
[tex]\boxed{ U_E = 9.16*10^{18}J/m^3}[/tex]
