Capacitor question by eureka forbes :)

See Electric field due to one plate on the other is
E = Q/2Aε
Therefore Force on the other plate will be Q²/2Aε
C = εA/d
Force = Q²/2Cd
Intial Force is F = Q²/2Cd
Q = CV given
f = C*V²/2d
Now The capacitor changes to C/1.5
Therefore Charge on the Capacitor is Q = C*2V/1.5
Force = Q²/2Cd this has been found before therefore
F= C*2V*2V/1.5*1.5*2d
F = 2*2/1.5*1.5 (CV²/2d)
F = 0.4*0.4/0.3*0.3f
F =16/9 (f).
It means The force becomes 16/9 times f