The electric potential at x = 8 m is −2000 V and at x = 2 m is +400 V. What is the magnitude and direction of the electric field?

Question

BlackTom AI · Accepted Answer

E = -dV/dx, so use the change in potential between x = 2 m and x = 8 m to find the (uniform) field.

1) Relevant concepts:
- Electric field and electric potential are related by E = -dV/dx (for one dimension). For a constant field between two points, E = -(V(8) − V(2))/(8 − 2).

2) Step-by-step reasoning:
- Take x1 = 2 m with V1 = +400 V and x2 = 8 m with V2 = −2000 V.
- Compute the potential difference ΔV = V2 − V1.
- Divide by the spatial separation Δx = x2 − x1 and apply the minus sign from E = -ΔV/Δx.

3) Calculations (all intermediate steps):
- ΔV = V(8) − V(2) = (−2000 V) − (+400 V) = −2400 V.
- Δx = 8 m − 2 m = 6 m.
- E = −(ΔV)/(Δx) = −(−2400 V)/(6 m) = +2400/6 V/m = +400 V/m.

4) Why this is correct:
- The positive sign of E means the field points in the +x direction. Physically, potential decreases with increasing x (from +400 V at x=2 to −2000 V at x=8), so the electric field must point in the +x direction (field points toward decreasing potential times a minus sign).

5) Why the other choices are incorrect:
- 300 V/m, 200 V/m, 600 V/m, and 500 V/m do not match the computed magnitude 2400/6 = 400 V/m; their magnitudes are inconsistent with the actual ΔV/Δx. Any answer with the wrong magnitude or wrong sign is therefore incorrect.

6) Conclusion:
The magnitude and direction are 400 V/m directed parallel to the +x-axis, which matches the provided correct answer.

The electric potential at x = 8 m is −2000 V and at x = 2 m is +400 V. What is the magnitude and direction of the electric field?单项选择题

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