A parallel plate capacitor of capacitance $$1\mu F$$ is charged to a potential difference of 20V. The distance between plates is $$1\mu m$$. The energy density between plates of capacitor is.

We need to find the energy density between the plates of a parallel plate capacitor.

The energy stored per unit volume (energy density) in an electric field is:

$$ u = \frac{1}{2}\epsilon_0 E^2 $$

This formula comes from the total energy stored in a capacitor $$U = \frac{1}{2}CV^2$$ divided by the volume between the plates.

For a parallel plate capacitor, the electric field is uniform and given by:

$$ E = \frac{V}{d} $$

where $$V = 20$$ V is the potential difference and $$d = 1\,\mu m = 1 \times 10^{-6}$$ m is the plate separation.

$$ E = \frac{20}{1 \times 10^{-6}} = 2 \times 10^7 \text{ V/m} $$

Using $$\epsilon_0 = 8.85 \times 10^{-12}$$ F/m (permittivity of free space):

$$ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (2 \times 10^7)^2 $$

$$ = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14} $$

$$ = \frac{1}{2} \times 8.85 \times 4 \times 10^{2} $$

$$ = \frac{1}{2} \times 35.4 \times 10^{2} $$

$$ = 17.7 \times 10^{2} \approx 1.77 \times 10^3 \approx 1.8 \times 10^3 \text{ J/m}^3 $$

The correct answer is Option C: $$1.8 \times 10^3$$ J/m$$^3$$.