What is the Initial Acceleration of a Charged Particle in an Electric Field?

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Homework Help Overview

The discussion revolves around calculating the initial acceleration of a charged particle in an electric field, specifically involving a particle with mass and charge interacting with another charged particle. The context includes the application of Coulomb's law and Newton's second law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the charged particle and the application of Coulomb's law to determine these forces. There are inquiries about how to relate the force to acceleration and the necessity of unit conversions, particularly from grams to kilograms.

Discussion Status

Participants are actively engaging with the problem, exploring the calculations involved in determining the force and its implications for acceleration. Guidance has been provided regarding the use of Coulomb's law and the importance of consistent units, but no consensus or resolution has been reached yet.

Contextual Notes

There is a focus on ensuring proper unit conversions and understanding the relationships between force, mass, and acceleration. The original poster's initial conditions and parameters are also noted, including the specific charges and distances involved.

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im stuck on this question if any1 could help i'd appreciate it. A particle of mass 30 g and charge 26 uC is released ffrom rest when it is 59 cm from a second particle of charge -11 uC. The question ask for the initial acceleration of the 30 g particle.
 
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To start with...

Try working out the forces being exerted by the particles.
 
What force acts on the particle? (Hint: Consider Coulomb's law)
 
so i work out the forces using coulombs law, but then how do u find acceleration in m/s?
 
Use Newton's 2nd law. (Acceleration has units of m/s^2, not m/s.)
 
heres what i got can u tell me if its on the right track

F = k (Q1) (Q2) / r^2

F = (9.0 x 10^9) (26uC) (-11uC) / (0.59m)^2


= -7.39 x 10^13 N
 
Last edited:
Realize that [itex]\mu \mbox{C}[/itex] means micro-Coulombs: 1 [itex]\mu \mbox{C} = 10^{-6} \mbox{C}[/itex].
 
when dividing by the mass do i have to change grams to kg?
 
Think of the units that make up a Newton and make sure you use the same
 
  • #10
thank u guys for ur help
 

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