Four metallic plates each with a surface area of one side A, are placed at a distance d from each other. The two outer plates are connected to one point A and the two other inner plates to another point B . Then the capacitance of the system is

 Applying Gauss’s law to this closed surface, we get,

∮E.ΔA=Qen/ε0=σΔA/ε0

⇒ EΔA =σΔA/ε0

⇒ E = σ/ε0 = Q/Aε0

Now, the potential difference between the plates will be V=V+−V−

=−∫E.dr.

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.