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Field and electron acceleration from potential

Short answer

The electric field components are E_x = −12 V/m, E_y = −27 V/m; magnitude |E| = √(144+729) ≈ 29.5 V/m. An electron experiences force opposite to E, so acceleration along +x is a_x = e|E_x|/m ≈ 2.11×10¹² m/s².

What is the magnitude of the electric field at the point (3,2) m if the electric potential is given by V = 2x + 5xy + 3y² volts. What acceleration does an electron experience in the x-direction?

Given V = 2x + 5xy + 3y² (volts), point (3, 2) m.

Field components

E = −∇V.

  1. Eₓ = −∂V/∂x = −(2 + 5y) = −(2 + 10) = −12 V/m
  2. E_y = −∂V/∂y = −(5x + 6y) = −(15 + 12) = −27 V/m

|E| = √(Eₓ² + E_y²) = √(144 + 729) = √873 = 29.5 V/m

Direction: tanφ = E_y/Eₓ = 27/12, so φ = 66.0° below the −x axis (components both negative).

Acceleration of an electron along x

Force on the electron: F = −eE, so Fₓ = −e Eₓ = −(1.6 × 10⁻¹⁹)(−12) = +1.92 × 10⁻¹⁸ N.

aₓ = Fₓ/m = 1.92 × 10⁻¹⁸ / 9.11 × 10⁻³¹ = 2.11 × 10¹² m/s²

Answer: |E| ≈ 29.5 V/m; the electron's acceleration in the x-direction is ≈ 2.11 × 10¹² m/s², along +x (opposite to Eₓ because the charge is negative).

✓ Answer checked by Claude Opus 5.5 (high effort)

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