A ball with a charge of +4.0 µC is carried slowly by an outside agent from a point A at 35 V to a…

Step 1. The work on the positive ball. $V_B - V_A = 155\ \text{V} - 35\ \text{V} = 120\ \text{V}$. So $W = q\,\Delta V = 4.0 \times 10^{-6}\ \text{C} \times 120\ \text{V} = 4.8 \times 10^{-4}\ \text{J}$, that is $0.48$ mJ.

Step 2. The release. The ball moves from B to A, from high to low potential. The potential energy falls by $0.48$ mJ, and the kinetic energy rises by the same amount. It starts from rest, so it has $0.48$ mJ at A.

Step 3. The negative ball. $W = q\,\Delta V = (-4.0 \times 10^{-6}\ \text{C}) \times 120\ \text{V} = -4.8 \times 10^{-4}\ \text{J}$, that is $-0.48$ mJ.