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Patdon10

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## 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?

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