CAPACITOR question with solution (Difficulty level : easy one)

Question-1:A Parallel Plate capacitor has following dimensions
Distance between the plates=10 cm
Area of Plate=2 m²
Charge on each plate=8.85 x 10^-10 C
Calculate following

1. Electric Field outside the plates
2. Electric Field Between the plates
3. Capacitance of the capacitor
4. Energy stored in the capacitor

ε₀=8.854 x 10^-12 C²N^-1m^-2
Solution-1:

As we know that Electric field outside the plates are zero
Electric field Inside the plates is
E=Q/ε₀A =50NC^-1

Capacitance=ε₀A/d =8.854 x 10^-12  x 2/.1=17.6 x 10^-11 F

Energy stored in capacitor=(1/2)Q²/C=.5 x 8.85 x 10^-10 x 8.85 x 10^-10/17.6 x 10^-11
=22.125 x 10^-10 J