Pls explain how we get the answer as clear as possible. Thx you so much ??

Chapter 21 - 11: Q : Consider an infinite flat sheet with positive charge density o in which a circular hole of radius R has been cut out . The sheet lies in the my-plane with the origin at the center of the hole The sheet is parallel to the ground, so that the positive 8-axis describes the "upward" direction . If a particle of mass in and negative charge -q sits at rest at the center of the hole and is released , the particle constrained to the z-axis , begins to fall . As it drops father beneath the sheet , the upward electric force increases . For a sufficiently low value of m , the upward electrical attraction eventually exceeds the particle's weight and the particle will slow , come to a stop , and then rise back to its original position . This sequence of events will repeat indefinitely. - Part A: What is the z-component of the electrical field at a depth A beneath the origin along the negative z- axis ? ANS . Ez ( A ) = - Part B: What is the maximum mass Mmmax that would prevent the particle from falling indefinitely ? ANS: mmax = - Part C: If me manx , how much work is done by the electric field as the particle drops from z=0 to 2=-A ? ANS: WE = 26. LR- JA'+ R= ] - Part D: In the interval of part, how much work is done by gravity ? ANS: Wg = mg4 - Part E: To what ultimate depth Amax will the particle drop ? ( C = m / m max ) ANS : Amax = - 2 RX 7 - 2 2 - Part F: What is the particle's speed at death As Amax ? ANS: IVIAll = / 294 - 22 5104 RX - R] - Part G: If the sheet has charge density 1.00n C/ am?, the radius of the hole R= 10.0 cm , and the particle has mass 31. 0g and charge - 1.00 UC , what is Mmax ? ANS : Mmax = 57.6g . - Part H: If the sheet has charge density 100 nC / am ' " the radius of the hole is R= 10.0cm , and the particle has mass 310g and change - 1.00 UC, what is Amax ? ANS: Amax = 15.2 cm