Given area of Parallel plate capacitor, A= 200 cm2 Separation between the plates, d=1.5 cm Force of attraction between the plates, F= 25×10−6 N F=QE F=2Aϵ0Q2 (E due to parallel plate =2ϵ0σ=A2ϵ0Q) But Q=CV=d∈0A(V) 
∴F=d2×2 Aϵ0(ϵ0AV2))=d2×2×(Aϵ0)(ϵ0A)2×V2=d2×2(ϵ0A)×V2 or, 25×10−6=2.25×10−4×2(8.85×10−12)×(200×10−4)×V2⇒V=8.85×10−12×200×10−425×10−6×2.25×10−4×2≈250 V