If moving from point A to B requires 3.5 μJ of work on a 14 nC charge and moving from C to B requires -5.0 μJ, what is VC − VA? (V = W/q)

• A. 607 V
• B. 250 V
• C. -607 V
• D. 0 V

Answer: (A)

The changing potential relates to the work done per unit charge: V_B − V_A = W_ext(A→B) / q. Here you do positive work 3.5 μJ on a 14 nC charge, so V_B − V_A = (3.5×10^−6 J) / (14×10^−9 C) = 250 V.

For movement from C to B, the external work is −5.0 μJ, so V_B − V_C = (−5.0×10^−6 J) / (14×10^−9 C) ≈ −357 V, which means V_C − V_B ≈ +357 V.

Now combine: V_C − V_A = (V_C − V_B) + (V_B − V_A) ≈ 357 V + 250 V = 607 V.