Acceleration of Charge in a Potential Gradient

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SUMMARY

The discussion centers on calculating the initial acceleration of a charged dust particle with a mass of 0.043g and a charge of 2.0 * 10^(-6)C in a potential gradient defined by the equation V(x) = (2.0V/m^2)x^2 - (3.0V/m^3)x^3. The particle is positioned at x = 2.1m. To find the acceleration, one must apply the equation for potential energy change, dU = qVba, and the relationship between potential and acceleration derived from Newton's second law.

PREREQUISITES
  • Understanding of electrostatics and potential energy
  • Familiarity with Newton's second law of motion
  • Knowledge of calculus for evaluating derivatives
  • Basic concepts of electric fields and forces
NEXT STEPS
  • Calculate the electric potential at x = 2.1m using V(x)
  • Determine the force acting on the particle using F = qE
  • Apply Newton's second law to find the acceleration, a = F/m
  • Explore the implications of varying potential gradients on particle motion
USEFUL FOR

Students studying physics, particularly those focusing on electrostatics and dynamics, as well as educators looking for examples of charge behavior in potential fields.

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Homework Statement



A dust particle with mass of 0.043g and a charge of 2.0 * 10^(-6)C is in a region of space where the potential is given by V(x) = (2.0V/m^2)x^2 - (3.0V/m^3)x^3. If the particle starts at x = 2.1m, what is the initial acceleration of the charge? Your explanation will be extremely helpful. Thank you for taking the time to look at this question.


Homework Equations



Vba = (Ub-Ua)/q
dU = qVba
Vba = Vb - Va

V = (1/(4pieo))*(Q/r)

The Attempt at a Solution



I don't even know where to begin with this question.
 
Last edited:
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Where is your attempt at solving the problem? What about relevant equations?
 

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