Solving for Acceleration of a Charged Particle

[Patdon10]

Homework Statement

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?

 

[Patdon10]

Apparently that is right, I just suck at algebra. (2.464*10^-5)/(2.5*10^-6) = 9.856 m/s^2, which is right.