1. The problem statement, all variables and given/known data A 2.50-mg dust particle with a charge of 2.00 µC falls at a point x = 2.80 m in a region where the electric potential varies according to V(x) = (2.00 V/m2)x2 − (1.00 V/m3)x3. With what acceleration will the particle start moving after it touches down? (Enter the magnitude of the acceleration.) 2. The attempt at a solution Taking the gradient of V will give you -E. So E = -4x + 3x^2 F = qE ---> F = (2e-6)(-4x+3x^2) F = (2e-6)(-11.2 + 23.52) F = 2.464e-5 We also know F = ma 2.464e-5 = (2.5e-6)a a = 6640.92 m/s^2 This is incorrect. Anyone know what I'm doing wrong?