Question

A dust particle with mass of 5.4×10−2 g and a charge of 2.3×10−6 C is in a region of space where the potential is given by V(x)=(1.3V/m2)x2−(2.7V/m3)x3. Part A If the particle starts at 2.3 m , what is the initial acceleration of the charge? Express your answer using two significant figures.

Answer 1

Answer:

1.6 m/s^2

Explanation:

Hello!

To calculate the acceleration we must know the electric field. The electric field and the potential are related by:

$E = -\frac{dV}{dx} =- 2.6(\frac{V}{m^{2}})x + 8.1(\frac{V}{m^{3}})x^{2}$

If the particle starts at 2.3m, the electric field is:

E = 36.869 V/m = 36.869 N/C

So, the force on the particle is:

F = q E =  2.3×10^−6 C * 36.869 N/C = 8.48 x 10^-5 N

And its acceleration is :

a = F/m =  8.48 x 10^-5 N / 5.4×10−5 kg = 1.57 m/s^2

Rounded to two significant figures:

1.6 m/s^2