Applying Gauss’s law to this closed surface, we get,
⇒ EΔA =σΔA/ε0
⇒ E = σ/ε0 = Q/Aε0
Now, the potential difference between the plates will be V=V+−V−
As you go from 2→1, the electric field and displacement are in opposite direction and therefore, E.dr = −Edr
V=−∫E.dr = ∫Edr = Ed =Qd/Aε0
Because capacitance is equal to QV, C=ε0A/d. Similarly, the capacitance of the other capacitor is C=ε0A/d. Because the positive and negative plates of both capacitors have similar points, the capacitances are in parallel. As a result, these two capacitors’ equivalent capacitance will be
C = C1 + C2 = ε0A/d + ε0A/d =2ε0A/d.Hence, the capacitance of the system is 2ε0Ad.