Answer:
[tex]7.6\times 10^{-6} J[/tex]
Explanation:
The electric field between two plates is:
[tex]E = \frac{\sigma}{2\epsilon}[/tex]
Where sigma is the superficial charge density and epsilon is the permittivity of air (is about the same as vaccum)
[tex]\sigma = \frac{Q}{Area}= \frac{530 \times 10^{-6} C}{(0.061m)^2} = 0.14C/m^2[/tex]
Using the formula to determine the electric field.
[tex]E = \frac{0.14C/m^2}{2\times 8.84 \times 10^{-12}C^2N^{-1}m^{-2}}\\E = 8.06 \times 10^9 N/C[/tex]
The energy inside the plates is given by the energy density due to electric Field times the volume:
[tex]Energy = \rho_{E} \times V = \frac{1}{2}\epsilon E V\\[/tex]
The volume is equal to the area of the plates times the distance between them:
[tex]Energy = \frac{1}{2}8.82\times 10^{-12}C^2 N^{-1}m^{-2} \times 8.06 \times 10^9 N/C \times (0.061m)^2\times 0.0035m\\Energy = 7.6 \times 10^{-6} J[/tex]
