Terminal velocity with equal drag forces

Steps:

• For rise at constant speed with V₁: q(V₁/d) = mg + F_d (electric up balances weight + drag down)
• For fall at constant speed with V₂: mg = q(V₂/d) + F_d (weight balances electric up + drag up)
• Add equations: q(V₁/d + V₂/d) = 2mg (drag terms cancel)
• Solve for charge: q = 2mg d / (V₁ + V₂)

Why D is correct:

• It derives from adding force balances, eliminating drag via superposition in opposite motions.

Why the others are wrong:

• A: Uses V₁ - V₂, which isolates 2F_d instead of charge.
• B: Lacks factor of 2, ignoring dual drag contributions.
• C: Combines factor of 2 with V₁ - V₂, yielding drag-related expression.

Final answer: D