Electric field intensity at the centre of the disc. E=2ϵ0σ (given) Electric field along the axis at any distance x from the centre of the disc E′=2ϵ0σ(1−x2+R2x) From question, x=R (radius of disc) ∴E′=2ϵ0σ(1−R2+R2R)=2ϵ0σ(2R2R−R) =144E ∴% reduction in the value of electric field =E(E−144E)×100=141000%=70.7%