1. Electric field between the plates
2.
Area under the straight line graph gives the energy stored in the capacitor.
3. Effective capacitance C = 20 × 5µF = 100µF, V = 100 v,
Energy stored E = \(\frac{1}{2}\)CV
E = \(\frac{1}{2}\times100\times10^{-6}\times100^2\) = 0.5 J