In a uniform electric field with potential at point A equal to -400 V and a field that moves a point to B located 8.5 cm to the right at 30 degrees above the horizontal, the field magnitude is 1.2×10^3 V/m. What is the potential at B?

• A. -488.32 V
• B. -400 V
• C. -500 V
• D. -450 V

Answer: (A)

When the electric field is uniform, the potential change between two points is given by the dot product with the displacement: ΔV = V_B − V_A = − E · Δr. Here the field is horizontal to the right with magnitude E = 1.2×10^3 V/m. The displacement from A to B has length 0.085 m and points 30° above the horizontal, so only the component along the field matters: E · Δr = E Δr cos(30°).

Compute the projection: Δr cos(30°) = 0.085 × 0.866025 ≈ 0.07361 m. Then E · Δr ≈ (1.2×10^3) × 0.07361 ≈ 88.33 V.

Thus V_B = V_A − E · Δr ≈ −400 V − 88.33 V ≈ −488.33 V, which matches −488.32 V when rounded.