(a) Deltav_(c)=(6)/(2)=3v across eachQ=C Delta V=(S_(mu)F)(BV)=15 mu C(b) c_(1)=kc=4cQ will remain same as disconneeted from battery.Q=C Delta V=C_(1)DeltaV_(1)^(')=>DeltaV_(1)^(')=(3)/(4)=0.75V{:[Deltav^(')=Deltav_(1)^(')+Deltav_(2)^(')=0.75+3],[Deltav^(')=3.75v]:}(c){:[w=Delta v=u_(f)-v_(i)=(1)/(2)(4xx5xx10^(-6))(0.75)^(2)-(1)/(2)(5xx10^(-6))(3)^(2)],[w=-16.875MJ]:}(d) finally, DeltaV_(1)^('')=Deltav_(2)^(n)=V_(f)Using cha ... See the full answer