can you please show how the calculation for this is done

Example 22.7 A proton and a charged sphere A proton is released from rest at the surface of a 1.0-cm-diameter sphere that has been charged to +1000 V. a. What is the charge of the sphere? b. What is the proton's speed at 1.0 cm from the sphere? PREPARE We'll assume that the motion takes place in a vacuum so that mechanical energy is conserved. The potential outside the charged sphere is the same as the potential of a point charge at the center. Figure 22.23IE| shows the situation. 1.0 cm SOLVE a. The charge of the sphere is Q = 42:49on0 = 0.56 x 109 c = 0.56 nC b. A sphere charged to V0 = +1000 V is positively charged. The proton will be repelled by this charge and move away from the sphere. The conservation of energy equation Kf + er = K, + eVi, with Equation 22.355I for the potential of a sphere, is 1 6R 1 3R v2 V = v.2 v 2mf+ rf 2m1+ r10 The proton starts from the surface of the sphere, ri = R, with vi = 0. When the proton is 1.0 cm from the surface of the sphere, it has rf = 1.0 cm + R = 1.5 cm. Using these, we can solve for Vf : vf='/2d (1R) =3.6x105m/s m rf