v0=RKQ=RKσ4πr2At center, v1−v0-(potential due to removed element)=v0−RKq=v0−RK(σα4πR2)=v0−v0α=v0(1−α)At mid point between center and hole is
V2=V0−R/2kq=V0−2V0α=V0(1−2α)V2V1=V0(1−2α)V0(1−α)=1−2α1−α→(2) is correct
Option A: Einitial=0Efinal=RzKq=R2kσα4πR2=RV0αEi−Ef=RV0α i.e E will decrease by RV0α i.e. (C) is wrongOption D:Vinitial=V0