A parallel plate capacitor with width 4cm , length 8cm and...

The figure below shows the parallel plate capacitor with dielectric slab placed in between the plates.

Now, the effective capacitance of the combination is

${C}_{eff} = \left[\dfrac{{ε}_{0}\left(7 \times 4\right)}{\dfrac{4}{10}} + \dfrac{5{ε}_{0}\left(1 \times 4\right)}{\dfrac{4}{10}}\right] \times {10}^{- 2}$

$= \left[70{ε}_{0} + 50{ε}_{0}\right] \times {10}^{- 2} = 120{ε}_{0} \times {10}^{- 2}$

$⇒ {C}_{eff} = 1 . 2{ε}_{0}$

Electrostatics energy of this system is $= \dfrac{1}{2}{C}_{eff}{V}^{2}$

$= \dfrac{1}{2}\left(1 . 2\right){ε}_{0}\left(20\right)\left(20\right) = 240{ε}_{0}$