Question

A $5.80\mu F$ parallel-plate air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 V. Calculate the energy density in the region between the plates, in $J/m^3$.

Answer 1

The expression of electrostatic energy density is $u_E=\frac{1}{2}{ϵ}_0{E}^2$

Here, $E=V/d=\frac{400}{5\times {10}^{-3}}=8\times {10}^{4}V/m$

Thus, $u_E=0.5\times (8.85\times {10}^{-12})\times (8\times {10}^4{)}^2=2.83\times {10}^{-2}J/m^3$