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Field from potential V = a + bx²

Short answer

Potentials: at x = 0 m, 10 V; at 3 m, −53 V; at 6 m, −242 V. E = −dV/dx = 14x; magnitude and direction: at 0 m, 0; at 3 m, 42 V/m (+x); at 6 m, 84 V/m (+x).

The potential in a region between x = 0m and x = 6m is V = a + bx² where a = 10 and b = −7V/m. Determine (i) the potentials at x = 0m, 3m and 6m and (ii) the magnitude and direction of electric fields at x = 0m, 3m and 6m.

Given V = a + bx² with a = 10 V and b = −7 V/m², so V = 10 − 7x². (b is given in V/m; dimensionally it is V/m².)

(i) Potentials

  • x = 0: V = 10 − 0 = 10 V
  • x = 3 m: V = 10 − 7(9) = 10 − 63 = −53 V
  • x = 6 m: V = 10 − 7(36) = 10 − 252 = −242 V

(ii) Electric fields

E = −dV/dx = −d(10 − 7x²)/dx = 14x (V/m)

x (m)E = 14x (V/m)Direction
00none
342+x
684+x

The field is positive, so it points along +x, the direction in which the potential decreases, as expected.

Answer: V = 10 V, −53 V, −242 V; E = 0, 42 V/m, 84 V/m (along +x) at x = 0, 3 m, 6 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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